Question

Solve each quadratic equation. Show your work. x^2 + 3x = 10

Answer 1

$x^{2} +3x=10$
$x^{2} +3x-10=0$
$x^{2} +5x-2x-10=0$
 x(x+5)-2(x+5)=0 
(x-2)(x+5)
x=2, x=-5

Answer 2

Answer:

Step-by-step explanation:

We shall use middle term split method to find the factor

here middle term is 3 ,factor of -10 which gives -3  on adding

such numbers are -10 = 5 and -2

                             5+(-2) = 3

                  And then making two groups and simplifying it ,we get the roots

$x^2+3x =10 \\x^2+3x-10 =0 \\x^2+(5-2)x-10=0 \\x^2+5x-2x-10=0 \\x(x+5)-2(x+5) = 0\\(x-2)(x+5) =0 \\x=2 or x= -5$