Question

What are the two solutions of 2x^2=-x^2-5x-1

Answer 1

$Solution, 2x^2=-x^2-5x-1: x=-\frac{5+\sqrt{13}}{6},\:x=-\frac{5-\sqrt{13}}{6}$

$Steps:$

$2x^2=-x^2-5x-1$

$\mathrm{Switch\:sides},\\-x^2-5x-1=2x^2$

$\mathrm{Subtract\:}2x^2\mathrm{\:from\:both\:sides},\\-x^2-5x-1-2x^2=2x^2-2x^2$

$\mathrm{Simplify},\\-3x^2-5x-1=0$

$Solve\:with\:the\:quadratic\:formula,\\\mathrm{For\:}\quad a=-3,\:b=-5,\:c=-1:\quad x_{1,\:2}=\frac{-\left(-5\right)\pm \sqrt{\left(-5\right)^2-4\left(-3\right)\left(-1\right)}}{2\left(-3\right)}\\x=\frac{-\left(-5\right)+\sqrt{\left(-5\right)^2-4\left(-3\right)\left(-1\right)}}{2\left(-3\right)}:\quad -\frac{5+\sqrt{13}}{6},\\x=\frac{-\left(-5\right)-\sqrt{\left(-5\right)^2-4\left(-3\right)\left(-1\right)}}{2\left(-3\right)}:\quad -\frac{5-\sqrt{13}}{6}$

$\mathrm{The\:final\:solutions\:to\:the\:quadratic\:equation\:are:}\\x=-\frac{5+\sqrt{13}}{6},\:x=-\frac{5-\sqrt{13}}{6}$

$Hope\:This\:Helps!!!$

$-Austint1414$

Answer 2

Answer:

x = 5/-6 + (√13)/-6

x = 5/-6 -(√13)/-6

Step-by-step explanation:

2x^2 = -x^2 - 5x - 1. Subtract 2x^2 from both sides.

-3x^2 - 5x - 1. Do the quadratic formula.

That gives you:

-5/6 ± (√13)/-6.