Question

2x²+ 3x - 720 solve the equation ​

Answer 1

$\huge\bold{\gray{\sf{Answer:}}}$

Explanation:

$2 {x}^{2} + 3x - 720$

It is in the form of a quadratic equation e.g $\bold{\red{a {x}^{2} + bx + c}}$Where ,c is a constant.

Now we factorise it to find roots

Note : It is a quadratic equation so it have maximum two zeroes or two roots.

=>A quadratic equation are those equation in which maximum power is 2.

How to factorise..??

step1: Multiply the constant term with coefficient of x square.suppose we obtain 'z' after multiplying

step 2:Think of a number whose sum makes middle term and product makes the 'z'

step 3: now split the middle term by that number which you think.

step 4:At last take common and compare with equal to 0 we obtain the roots .

Now come to the question:-

$2 {x}^{2} + 3x - 720$

$\bold{720\times2=1440}$Now,we have to think of a number whose sum makes 3 and product makes 1440.

Here ,we use quadratic formula to find out the roots because we are unable to find any number whose sum is 3 and product is 1440.

Here,a=2,b=3 and c=-720

$x = \frac{ - b ± \sqrt{ {b}^{2} - 4ac} }{2ac}$

$= > x = \frac{ - 3 ± \sqrt{ {(3)}^{2} - 4 \times 2 \times - 720} }{2 \times 2 - 720}$

$= > x = \frac{ - 3 ± \sqrt{9 + 5760} }{ - 2880}$

$= > x = \frac{ - 3 ± \sqrt{5769} }{ - 2880}$

$= > x = \frac{ - 3 ± 3 \sqrt{641} }{ - 2880}$

$= > x = \frac{3(1 ± \sqrt{641)} }{2880}$

$\bold{x = \frac{1 ± \sqrt{641} }{960}}$

Answer 2

$\huge\bold{\gray{\sf{Answer:}}}$

Explanation:

$2 {x}^{2} + 3x - 720$

It is in the form of a quadratic equation e.g $\bold{\red{a {x}^{2} + bx + c}}$Where ,c is a constant.

Now we factorise it to find roots

Note : It is a quadratic equation so it have maximum two zeroes or two roots.

=>A quadratic equation are those equation in which maximum power is 2.

How to factorise..??

step1: Multiply the constant term with coefficient of x square.suppose we obtain 'z' after multiplying

step 2:Think of a number whose sum makes middle term and product makes the 'z'

step 3: now split the middle term by that number which you think.

step 4:At last take common and compare with equal to 0 we obtain the roots .

Now come to the question:-

$2 {x}^{2} + 3x - 720$

$\bold{720\times2=1440}$Now,we have to think of a number whose sum makes 3 and product makes 1440.

Here ,we use quadratic formula to find out the roots because we are unable to find any number whose sum is 3 and product is 1440.

Here,a=2,b=3 and c=-720

$x = \frac{ - b ± \sqrt{ {b}^{2} - 4ac} }{2ac}$

$= > x = \frac{ - 3 ± \sqrt{ {(3)}^{2} - 4 \times 2 \times - 720} }{2 \times 2 - 720}$

$= > x = \frac{ - 3 ± \sqrt{9 + 5760} }{ - 2880}$

$= > x = \frac{ - 3 ± \sqrt{5769} }{ - 2880}$

$= > x = \frac{ - 3 ± 3 \sqrt{641} }{ - 2880}$

$= > x = \frac{3(1 ± \sqrt{641)} }{2880}$

$\bold{x = \frac{1 ± \sqrt{641} }{960}}$