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x^{2}-290x+18000=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{-\left(-290\right)±\sqrt{\left(-290\right)^{2}-4\times 18000}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -290 for b, and 18000 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-290\right)±\sqrt{84100-4\times 18000}}{2}
Square -290.
x=\frac{-\left(-290\right)±\sqrt{84100-72000}}{2}
Multiply -4 times 18000.
x=\frac{-\left(-290\right)±\sqrt{12100}}{2}
Add 84100 to -72000.
x=\frac{-\left(-290\right)±110}{2}
Take the square root of 12100.
x=\frac{290±110}{2}
The opposite of -290 is 290.
x=\frac{400}{2}
Now solve the equation x=\frac{290±110}{2} when ± is plus. Add 290 to 110.
x=200
Divide 400 by 2.
x=\frac{180}{2}
Now solve the equation x=\frac{290±110}{2} when ± is minus. Subtract 110 from 290.
x=90
Divide 180 by 2.
x=200 x=90
The equation is now solved.
x^{2}-290x+18000=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}-290x+18000-18000=-18000
Subtract 18000 from both sides of the equation.
x^{2}-290x=-18000
Subtracting 18000 from itself leaves 0.
x^{2}-290x+\left(-145\right)^{2}=-18000+\left(-145\right)^{2}
Divide -290, the coefficient of the x term, by 2 to get -145. Then add the square of -145 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-290x+21025=-18000+21025
Square -145.
x^{2}-290x+21025=3025
Add -18000 to 21025.
\left(x-145\right)^{2}=3025
Factor x^{2}-290x+21025. 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-145\right)^{2}}=\sqrt{3025}
Take the square root of both sides of the equation.
x-145=55 x-145=-55
Simplify.
x=200 x=90
Add 145 to both sides of the equation.