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2x^{2}+5x=6
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.
2x^{2}+5x-6=6-6
Subtract 6 from both sides of the equation.
2x^{2}+5x-6=0
Subtracting 6 from itself leaves 0.
x=\frac{-5±\sqrt{5^{2}-4\times 2\left(-6\right)}}{2\times 2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 5 for b, and -6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-5±\sqrt{25-4\times 2\left(-6\right)}}{2\times 2}
Square 5.
x=\frac{-5±\sqrt{25-8\left(-6\right)}}{2\times 2}
Multiply -4 times 2.
x=\frac{-5±\sqrt{25+48}}{2\times 2}
Multiply -8 times -6.
x=\frac{-5±\sqrt{73}}{2\times 2}
Add 25 to 48.
x=\frac{-5±\sqrt{73}}{4}
Multiply 2 times 2.
x=\frac{\sqrt{73}-5}{4}
Now solve the equation x=\frac{-5±\sqrt{73}}{4} when ± is plus. Add -5 to \sqrt{73}.
x=\frac{-\sqrt{73}-5}{4}
Now solve the equation x=\frac{-5±\sqrt{73}}{4} when ± is minus. Subtract \sqrt{73} from -5.
x=\frac{\sqrt{73}-5}{4} x=\frac{-\sqrt{73}-5}{4}
The equation is now solved.
2x^{2}+5x=6
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.
\frac{2x^{2}+5x}{2}=\frac{6}{2}
Divide both sides by 2.
x^{2}+\frac{5}{2}x=\frac{6}{2}
Dividing by 2 undoes the multiplication by 2.
x^{2}+\frac{5}{2}x=3
Divide 6 by 2.
x^{2}+\frac{5}{2}x+\left(\frac{5}{4}\right)^{2}=3+\left(\frac{5}{4}\right)^{2}
Divide \frac{5}{2}, the coefficient of the x term, by 2 to get \frac{5}{4}. Then add the square of \frac{5}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+\frac{5}{2}x+\frac{25}{16}=3+\frac{25}{16}
Square \frac{5}{4} by squaring both the numerator and the denominator of the fraction.
x^{2}+\frac{5}{2}x+\frac{25}{16}=\frac{73}{16}
Add 3 to \frac{25}{16}.
\left(x+\frac{5}{4}\right)^{2}=\frac{73}{16}
Factor x^{2}+\frac{5}{2}x+\frac{25}{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+\frac{5}{4}\right)^{2}}=\sqrt{\frac{73}{16}}
Take the square root of both sides of the equation.
x+\frac{5}{4}=\frac{\sqrt{73}}{4} x+\frac{5}{4}=-\frac{\sqrt{73}}{4}
Simplify.
x=\frac{\sqrt{73}-5}{4} x=\frac{-\sqrt{73}-5}{4}
Subtract \frac{5}{4} from both sides of the equation.