Solve -x^2-5=-6x^2

$$- x ^ { 2 } - 5 = - 6 x ^ { 2 }$$

Solve for x

$$x=1$$

$$x=-1$$

Steps Using Difference of Squares

Add 6x^{2} to both sides.

$$-x^{2}-5+6x^{2}=0$$

Combine -x^{2} and 6x^{2} to get 5x^{2}.

$$5x^{2}-5=0$$

Divide both sides by 5.

$$x^{2}-1=0$$

Consider x^{2}-1. Rewrite x^{2}-1 as x^{2}-1^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(x-1\right)\left(x+1\right)=0$$

To find equation solutions, solve x-1=0 and x+1=0.

$$x=1 x=-1$$

Steps by Finding Square Root

Add 6x^{2} to both sides.

$$-x^{2}-5+6x^{2}=0$$

Add 5 to both sides. Anything plus zero gives itself.

$$-x^{2}+6x^{2}=5$$

Combine -x^{2} and 6x^{2} to get 5x^{2}.

$$5x^{2}=5$$

Divide both sides by 5.

$$x^{2}=\frac{5}{5}$$

Divide 5 by 5 to get 1.

$$x^{2}=1$$

Take the square root of both sides of the equation.

$$x=1 x=-1$$

Steps Using the Quadratic Formula

Add 6x^{2} to both sides.

$$-x^{2}-5+6x^{2}=0$$

Combine -x^{2} and 6x^{2} to get 5x^{2}.

$$5x^{2}-5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, 0 for b, and -5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\times 5\left(-5\right)}}{2\times 5}$$

Square 0.

$$x=\frac{0±\sqrt{-4\times 5\left(-5\right)}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{0±\sqrt{-20\left(-5\right)}}{2\times 5}$$

Multiply -20 times -5.

$$x=\frac{0±\sqrt{100}}{2\times 5}$$

Take the square root of 100.

$$x=\frac{0±10}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{0±10}{10}$$

Now solve the equation x=\frac{0±10}{10} when ± is plus. Divide 10 by 10.

$$x=1$$

Now solve the equation x=\frac{0±10}{10} when ± is minus. Divide -10 by 10.

$$x=-1$$

The equation is now solved.

$$x=1 x=-1$$