Solve -5+x^2div4=11

$$- 5 + x ^ { 2 } \div 4 = 11$$

Solve for x

$$x=8$$

$$x=-8$$

Steps Using Difference of Squares

Multiply both sides of the equation by 4.

$$-20+x^{2}=44$$

Subtract 44 from both sides.

$$-20+x^{2}-44=0$$

Subtract 44 from -20 to get -64.

$$-64+x^{2}=0$$

Consider -64+x^{2}. Rewrite -64+x^{2} as x^{2}-8^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(x-8\right)\left(x+8\right)=0$$

To find equation solutions, solve x-8=0 and x+8=0.

$$x=8 x=-8$$

Steps by Finding Square Root

Multiply both sides of the equation by 4.

$$-20+x^{2}=44$$

Add 20 to both sides.

$$x^{2}=44+20$$

Add 44 and 20 to get 64.

$$x^{2}=64$$

Take the square root of both sides of the equation.

$$x=8 x=-8$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 4.

$$-20+x^{2}=44$$

Subtract 44 from both sides.

$$-20+x^{2}-44=0$$

Subtract 44 from -20 to get -64.

$$-64+x^{2}=0$$

Quadratic equations like this one, with an x^{2} term but no x term, can still be solved using the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}, once they are put in standard form: ax^{2}+bx+c=0.

$$x^{2}-64=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 0 for b, and -64 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\left(-64\right)}}{2}$$

Square 0.

$$x=\frac{0±\sqrt{-4\left(-64\right)}}{2}$$

Multiply -4 times -64.

$$x=\frac{0±\sqrt{256}}{2}$$

Take the square root of 256.

$$x=\frac{0±16}{2}$$

Now solve the equation x=\frac{0±16}{2} when ± is plus. Divide 16 by 2.

$$x=8$$

Now solve the equation x=\frac{0±16}{2} when ± is minus. Divide -16 by 2.

$$x=-8$$

The equation is now solved.

$$x=8 x=-8$$