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5x^{2}+3x-6=25
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.
5x^{2}+3x-6-25=25-25
Subtract 25 from both sides of the equation.
5x^{2}+3x-6-25=0
Subtracting 25 from itself leaves 0.
5x^{2}+3x-31=0
Subtract 25 from -6.
x=\frac{-3±\sqrt{3^{2}-4\times 5\left(-31\right)}}{2\times 5}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, 3 for b, and -31 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-3±\sqrt{9-4\times 5\left(-31\right)}}{2\times 5}
Square 3.
x=\frac{-3±\sqrt{9-20\left(-31\right)}}{2\times 5}
Multiply -4 times 5.
x=\frac{-3±\sqrt{9+620}}{2\times 5}
Multiply -20 times -31.
x=\frac{-3±\sqrt{629}}{2\times 5}
Add 9 to 620.
x=\frac{-3±\sqrt{629}}{10}
Multiply 2 times 5.
x=\frac{\sqrt{629}-3}{10}
Now solve the equation x=\frac{-3±\sqrt{629}}{10} when ± is plus. Add -3 to \sqrt{629}.
x=\frac{-\sqrt{629}-3}{10}
Now solve the equation x=\frac{-3±\sqrt{629}}{10} when ± is minus. Subtract \sqrt{629} from -3.
x=\frac{\sqrt{629}-3}{10} x=\frac{-\sqrt{629}-3}{10}
The equation is now solved.
5x^{2}+3x-6=25
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.
5x^{2}+3x-6-\left(-6\right)=25-\left(-6\right)
Add 6 to both sides of the equation.
5x^{2}+3x=25-\left(-6\right)
Subtracting -6 from itself leaves 0.
5x^{2}+3x=31
Subtract -6 from 25.
\frac{5x^{2}+3x}{5}=\frac{31}{5}
Divide both sides by 5.
x^{2}+\frac{3}{5}x=\frac{31}{5}
Dividing by 5 undoes the multiplication by 5.
x^{2}+\frac{3}{5}x+\left(\frac{3}{10}\right)^{2}=\frac{31}{5}+\left(\frac{3}{10}\right)^{2}
Divide \frac{3}{5}, the coefficient of the x term, by 2 to get \frac{3}{10}. Then add the square of \frac{3}{10} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+\frac{3}{5}x+\frac{9}{100}=\frac{31}{5}+\frac{9}{100}
Square \frac{3}{10} by squaring both the numerator and the denominator of the fraction.
x^{2}+\frac{3}{5}x+\frac{9}{100}=\frac{629}{100}
Add \frac{31}{5} to \frac{9}{100} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x+\frac{3}{10}\right)^{2}=\frac{629}{100}
Factor x^{2}+\frac{3}{5}x+\frac{9}{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}.
\sqrt{\left(x+\frac{3}{10}\right)^{2}}=\sqrt{\frac{629}{100}}
Take the square root of both sides of the equation.
x+\frac{3}{10}=\frac{\sqrt{629}}{10} x+\frac{3}{10}=-\frac{\sqrt{629}}{10}
Simplify.
x=\frac{\sqrt{629}-3}{10} x=\frac{-\sqrt{629}-3}{10}
Subtract \frac{3}{10} from both sides of the equation.