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x^{2}+18x-95=0
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=\frac{-18±\sqrt{18^{2}-4\left(-95\right)}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 18 for b, and -95 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-18±\sqrt{324-4\left(-95\right)}}{2}
Square 18.
x=\frac{-18±\sqrt{324+380}}{2}
Multiply -4 times -95.
x=\frac{-18±\sqrt{704}}{2}
Add 324 to 380.
x=\frac{-18±8\sqrt{11}}{2}
Take the square root of 704.
x=\frac{8\sqrt{11}-18}{2}
Now solve the equation x=\frac{-18±8\sqrt{11}}{2} when ± is plus. Add -18 to 8\sqrt{11}.
x=4\sqrt{11}-9
Divide -18+8\sqrt{11} by 2.
x=\frac{-8\sqrt{11}-18}{2}
Now solve the equation x=\frac{-18±8\sqrt{11}}{2} when ± is minus. Subtract 8\sqrt{11} from -18.
x=-4\sqrt{11}-9
Divide -18-8\sqrt{11} by 2.
x=4\sqrt{11}-9 x=-4\sqrt{11}-9
The equation is now solved.
x^{2}+18x-95=0
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}+18x-95-\left(-95\right)=-\left(-95\right)
Add 95 to both sides of the equation.
x^{2}+18x=-\left(-95\right)
Subtracting -95 from itself leaves 0.
x^{2}+18x=95
Subtract -95 from 0.
x^{2}+18x+9^{2}=95+9^{2}
Divide 18, the coefficient of the x term, by 2 to get 9. Then add the square of 9 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+18x+81=95+81
Square 9.
x^{2}+18x+81=176
Add 95 to 81.
\left(x+9\right)^{2}=176
Factor x^{2}+18x+81. 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+9\right)^{2}}=\sqrt{176}
Take the square root of both sides of the equation.
x+9=4\sqrt{11} x+9=-4\sqrt{11}
Simplify.
x=4\sqrt{11}-9 x=-4\sqrt{11}-9
Subtract 9 from both sides of the equation.