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x^{2}+8x=38
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}+8x-38=38-38
Subtract 38 from both sides of the equation.
x^{2}+8x-38=0
Subtracting 38 from itself leaves 0.
x=\frac{-8±\sqrt{8^{2}-4\left(-38\right)}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 8 for b, and -38 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-8±\sqrt{64-4\left(-38\right)}}{2}
Square 8.
x=\frac{-8±\sqrt{64+152}}{2}
Multiply -4 times -38.
x=\frac{-8±\sqrt{216}}{2}
Add 64 to 152.
x=\frac{-8±6\sqrt{6}}{2}
Take the square root of 216.
x=\frac{6\sqrt{6}-8}{2}
Now solve the equation x=\frac{-8±6\sqrt{6}}{2} when ± is plus. Add -8 to 6\sqrt{6}.
x=3\sqrt{6}-4
Divide 6\sqrt{6}-8 by 2.
x=\frac{-6\sqrt{6}-8}{2}
Now solve the equation x=\frac{-8±6\sqrt{6}}{2} when ± is minus. Subtract 6\sqrt{6} from -8.
x=-3\sqrt{6}-4
Divide -8-6\sqrt{6} by 2.
x=3\sqrt{6}-4 x=-3\sqrt{6}-4
The equation is now solved.
x^{2}+8x=38
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}+8x+4^{2}=38+4^{2}
Divide 8, the coefficient of the x term, by 2 to get 4. Then add the square of 4 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+8x+16=38+16
Square 4.
x^{2}+8x+16=54
Add 38 to 16.
\left(x+4\right)^{2}=54
Factor x^{2}+8x+16. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.
\sqrt{\left(x+4\right)^{2}}=\sqrt{54}
Take the square root of both sides of the equation.
x+4=3\sqrt{6} x+4=-3\sqrt{6}
Simplify.
x=3\sqrt{6}-4 x=-3\sqrt{6}-4
Subtract 4 from both sides of the equation.