Question

2x^2-5x-3=0 and next question 7m^2-84=0

​

Answer 1

SOLUTION:

2x^2 - 5x - 3 = 0

2x^2 - (6 - 1)x - 3 = 0

2x^2 - 6x - x - 3 = 0

2x (x -3) - 1 (x - 3) = 0

(x - 3) (2x - 1) = 0

Now,

x - 3 = 0

x = 3

And,

(2x - 1) = 0

2x = 1

x = 1/2

Hence, 3 and 1/2 are the values of x

NEXT QUESTION:

7m^2 - 84 = 0

7m^2 = 84

m^2 = 84/7

m^2 = 12

$m \: = \sqrt{12}$

$m = \sqrt[2]{3}$

Answer 2

Answer:

$\huge{\red{\boxed{\boxed{\green{\sf{x=\frac{-1}{2},3}}}}}}$

$\huge{\red{\boxed{\boxed{\green{\sf{m=-\sqrt{12},\sqrt{12}}}}}}}$

Step-by-step explanation:

$\huge{\red{\boxed{\boxed{\green{\underline{\red{\sf{SOLUTION-}}}}}}}}$

☆First question:

□2x^2-5x-3=0

☆Second question:

□7m^2-84=0

To find:

○values of x=?

○values of m=?

$1st \: question \\ {2x}^{2} - 5x - 3= 0 \\ Quadratic \: formula\\ = ) d= b^{2} - 4ac\\ = )d=(-5)^{2}-4(2\times-3) \\ = )d=25+24 \\ =)d= 49\\\\=) x = \frac{-b+\sqrt{d}}{2a}\\=)x=\frac{5+7}{4}\\ = )x = 3 - - - - - 1st \: value \\ \\ =)x =\frac{-b-\sqrt{d}}{2a} \\ = )x = \frac{5-7}{2\times2}\\=)x=\frac{-2}{4} \\=)x=\frac{-1}{2}- - - - - 2nd \: value \\ \\ 2 nd \: question \\ {7m}^{2} - 84 = 0 \\ take \: common \: from \: these \: two \: values \\ = )7( {m}^{2} - 12) = 0 \\ = ) {m}^{2} - 12 = 0 \\ = ) {m}^{2} = 12\\ = )m = + - \sqrt{12} \\ \\ = )m = - \sqrt{12} - - - - - 1st \: values \\ \\ = )m = \sqrt{12} - - - - - 2nd \: value$

$\huge{\red{\boxed{\boxed{\green{\sf{x=\frac{-1}{2},3}}}}}}$

$\huge{\red{\boxed{\boxed{\green{\sf{m=-\sqrt{12},\sqrt{12}}}}}}}$

_________________________________________