Question

Solve 2x2 + 20x = −38. (1 point)

Answer 1

Answer:

The solutions are $x=-5+\sqrt{6}$  and $x=-5-\sqrt{6}$

Step-by-step 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

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

$x=\frac{-10(+/-)\sqrt{100-76}}{2}$

$x=\frac{-10(+/-)\sqrt{24}}{2}$

$x=\frac{-10(+/-)2\sqrt{6}}{2}$

$x1=\frac{-10(+)2\sqrt{6}}{2}=-5+\sqrt{6}$

$x2=\frac{-10(-)2\sqrt{6}}{2}=-5-\sqrt{6}$

Answer 2

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