Question

factorise the following question

1). 6x^2+10x-35​

Answer 1

factorise the following question

1). 6x²+10x-35

• By completing square method

$a = > 6 \\ \\ b = > 10 \\ \\ c = > - 35$

$\boxed{ \underline \bold \pink{d = > {b}^{2} - 4ac}}$

$d = > {(10)}^{2} - 4 \times 6 \times - 35 \\ \\ d = > 100 + 840 \\ \\ \boxed{d = > 940}$

$\boxed{ \underline \bold \pink{x = \frac{ - b + - \sqrt{d} }{2a} }}$

$x = > \frac{ - 10 + - \sqrt{940} }{2 \times 6} \\ \\ x = > \frac{ - 10 + \sqrt{940} }{2 \times 6} \: \: \: \: \: \: \: \: x = > \frac{ - 10 - \sqrt{940} }{2 \times 6} \\ \\ x = > \frac{ - 10 + \sqrt{940} }{12} \: \: \: \: \: \: \: \: \: \: \: x = > \frac{ - 10 - \sqrt{940} }{12}$

Hope it's helps you ☺️

Answer 2

The roots of $6x^2 +10x-35$ are $x= \frac{-10+\sqrt{940} }{12}$  and $x= \frac{-10-\sqrt{940} }{12}$

Step-by-step explanation:

given :

$6x^2 +10x-35$

using the quadratic formula to factorize

here ,

a= 6

b=10

c = - 35

then, $d= b^2-4ac$

$d= (10)^2-4\times6\times (-35)$

$d= 100+840$

$d= 940$

if $d< 0$ then the equation will have no real roots

here  $d> 0$ ( $940> 0$)

then ,

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

$x= \frac{-10\pm\sqrt{940} }{2\times 6}$

$x= \frac{-10\pm\sqrt{940} }{12}$

$x= \frac{-10+\sqrt{940} }{12}$  ,  $x= \frac{-10-\sqrt{940} }{12}$

hence ,

The roots of $6x^2 +10x-35$ are  $x= \frac{-10+\sqrt{940} }{12}$ and  $x= \frac{-10-\sqrt{940} }{12}$

#Learn more:

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