Solve x^2+10x+25=27

$$x ^ { 2 } + 10 x + 25 = 27$$

Solve for x

$$x=3\sqrt{3}-5\approx 0.196152423$$

$$x=-3\sqrt{3}-5\approx -10.196152423$$

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.

$$x^{2}+10x+25=27$$

Subtract 27 from both sides of the equation.

$$x^{2}+10x+25-27=27-27$$

Subtracting 27 from itself leaves 0.

$$x^{2}+10x+25-27=0$$

Subtract 27 from 25.

$$x^{2}+10x-2=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 10 for b, and -2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-10±\sqrt{10^{2}-4\left(-2\right)}}{2}$$

Square 10.

$$x=\frac{-10±\sqrt{100-4\left(-2\right)}}{2}$$

Multiply -4 times -2.

$$x=\frac{-10±\sqrt{100+8}}{2}$$

Add 100 to 8.

$$x=\frac{-10±\sqrt{108}}{2}$$

Take the square root of 108.

$$x=\frac{-10±6\sqrt{3}}{2}$$

Now solve the equation x=\frac{-10±6\sqrt{3}}{2} when ± is plus. Add -10 to 6\sqrt{3}.

$$x=\frac{6\sqrt{3}-10}{2}$$

Divide -10+6\sqrt{3} by 2.

$$x=3\sqrt{3}-5$$

Now solve the equation x=\frac{-10±6\sqrt{3}}{2} when ± is minus. Subtract 6\sqrt{3} from -10.

$$x=\frac{-6\sqrt{3}-10}{2}$$

Divide -10-6\sqrt{3} by 2.

$$x=-3\sqrt{3}-5$$

The equation is now solved.

$$x=3\sqrt{3}-5 x=-3\sqrt{3}-5$$

Steps for Completing the Square

Factor x^{2}+10x+25. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+5\right)^{2}=27$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+5\right)^{2}}=\sqrt{27}$$

Simplify.

$$x+5=3\sqrt{3} x+5=-3\sqrt{3}$$

Subtract 5 from both sides of the equation.

$$x=3\sqrt{3}-5 x=-3\sqrt{3}-5$$