Question

Solve 2x2 + 20x = −38. (1 point)

Answer 1

`Pull \ out \ like \ factors : \\ \\ 42 + 20x = 2 * (10x + 21) \ \\ \\ Solve : \ 2 = \ 0 \\ \\ Subtract 21 from both sides of the equation : \\ 10x = -21 \\ \\ Divide both sides of the equation by 10: \\ \\ x = -21/10 = \boxed{-2.100 }`

Answer 2

The solutions are
`x=-5+√(6)` and
`x=-5-√(6)`

Explanation:

we have

`2x^(2) +20x=-38`

Divide by
`2` both sides

`x^(2) +10x=-19` ------>
`x^(2) +10x+19=0`

we know that

The formula to solve a quadratic equation of the form
`ax^(2) +bx+c=0` is equal to

`x=\frac{-b(+/-)\sqrt{b^(2)-4ac}} {2a}`

in this problem we have

`x^(2) +10x+19=0`

so

`a=1\\b=10\\c=19`

substitute

`x=\frac{-10(+/-)\sqrt{10^(2)-4(1)(19)}}{2(1)}`

`x=(-10(+/-)√(100-76))/(2)`

`x=(-10(+/-)√(24))/(2)`

`x=(-10(+/-)2√(6))/(2)`

`x1=(-10(+)2√(6))/(2)=-5+√(6)`

`x2=(-10(-)2√(6))/(2)=-5-√(6)`