Solve x^2+20x+26=50

$$x ^ { 2 } + 20 x + 26 = 50$$

Solve for x

$$x=2\sqrt{31}-10\approx 1.135528726$$

$$x=-2\sqrt{31}-10\approx -21.135528726$$

Steps Using the Quadratic Formula

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$x^{2}+20x+26=50$$

Subtract 50 from both sides of the equation.

$$x^{2}+20x+26-50=50-50$$

Subtracting 50 from itself leaves 0.

$$x^{2}+20x+26-50=0$$

Subtract 50 from 26.

$$x^{2}+20x-24=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 20 for b, and -24 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-20±\sqrt{20^{2}-4\left(-24\right)}}{2}$$

Square 20.

$$x=\frac{-20±\sqrt{400-4\left(-24\right)}}{2}$$

Multiply -4 times -24.

$$x=\frac{-20±\sqrt{400+96}}{2}$$

Add 400 to 96.

$$x=\frac{-20±\sqrt{496}}{2}$$

Take the square root of 496.

$$x=\frac{-20±4\sqrt{31}}{2}$$

Now solve the equation x=\frac{-20±4\sqrt{31}}{2} when ± is plus. Add -20 to 4\sqrt{31}.

$$x=\frac{4\sqrt{31}-20}{2}$$

Divide -20+4\sqrt{31} by 2.

$$x=2\sqrt{31}-10$$

Now solve the equation x=\frac{-20±4\sqrt{31}}{2} when ± is minus. Subtract 4\sqrt{31} from -20.

$$x=\frac{-4\sqrt{31}-20}{2}$$

Divide -20-4\sqrt{31} by 2.

$$x=-2\sqrt{31}-10$$

The equation is now solved.

$$x=2\sqrt{31}-10 x=-2\sqrt{31}-10$$

Steps for Completing the Square

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$x^{2}+20x+26=50$$

Subtract 26 from both sides of the equation.

$$x^{2}+20x+26-26=50-26$$

Subtracting 26 from itself leaves 0.

$$x^{2}+20x=50-26$$

Subtract 26 from 50.

$$x^{2}+20x=24$$

Divide 20, the coefficient of the x term, by 2 to get 10. Then add the square of 10 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+20x+10^{2}=24+10^{2}$$

Square 10.

$$x^{2}+20x+100=24+100$$

Add 24 to 100.

$$x^{2}+20x+100=124$$

Factor x^{2}+20x+100. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+10\right)^{2}=124$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+10\right)^{2}}=\sqrt{124}$$

Simplify.

$$x+10=2\sqrt{31} x+10=-2\sqrt{31}$$

Subtract 10 from both sides of the equation.

$$x=2\sqrt{31}-10 x=-2\sqrt{31}-10$$