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4x^{2}+5x-2=30
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.
4x^{2}+5x-2-30=30-30
Subtract 30 from both sides of the equation.
4x^{2}+5x-2-30=0
Subtracting 30 from itself leaves 0.
4x^{2}+5x-32=0
Subtract 30 from -2.
x=\frac{-5±\sqrt{5^{2}-4\times 4\left(-32\right)}}{2\times 4}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, 5 for b, and -32 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-5±\sqrt{25-4\times 4\left(-32\right)}}{2\times 4}
Square 5.
x=\frac{-5±\sqrt{25-16\left(-32\right)}}{2\times 4}
Multiply -4 times 4.
x=\frac{-5±\sqrt{25+512}}{2\times 4}
Multiply -16 times -32.
x=\frac{-5±\sqrt{537}}{2\times 4}
Add 25 to 512.
x=\frac{-5±\sqrt{537}}{8}
Multiply 2 times 4.
x=\frac{\sqrt{537}-5}{8}
Now solve the equation x=\frac{-5±\sqrt{537}}{8} when ± is plus. Add -5 to \sqrt{537}.
x=\frac{-\sqrt{537}-5}{8}
Now solve the equation x=\frac{-5±\sqrt{537}}{8} when ± is minus. Subtract \sqrt{537} from -5.
x=\frac{\sqrt{537}-5}{8} x=\frac{-\sqrt{537}-5}{8}
The equation is now solved.
4x^{2}+5x-2=30
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.
4x^{2}+5x-2-\left(-2\right)=30-\left(-2\right)
Add 2 to both sides of the equation.
4x^{2}+5x=30-\left(-2\right)
Subtracting -2 from itself leaves 0.
4x^{2}+5x=32
Subtract -2 from 30.
\frac{4x^{2}+5x}{4}=\frac{32}{4}
Divide both sides by 4.
x^{2}+\frac{5}{4}x=\frac{32}{4}
Dividing by 4 undoes the multiplication by 4.
x^{2}+\frac{5}{4}x=8
Divide 32 by 4.
x^{2}+\frac{5}{4}x+\left(\frac{5}{8}\right)^{2}=8+\left(\frac{5}{8}\right)^{2}
Divide \frac{5}{4}, the coefficient of the x term, by 2 to get \frac{5}{8}. Then add the square of \frac{5}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+\frac{5}{4}x+\frac{25}{64}=8+\frac{25}{64}
Square \frac{5}{8} by squaring both the numerator and the denominator of the fraction.
x^{2}+\frac{5}{4}x+\frac{25}{64}=\frac{537}{64}
Add 8 to \frac{25}{64}.
\left(x+\frac{5}{8}\right)^{2}=\frac{537}{64}
Factor x^{2}+\frac{5}{4}x+\frac{25}{64}. 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}{8}\right)^{2}}=\sqrt{\frac{537}{64}}
Take the square root of both sides of the equation.
x+\frac{5}{8}=\frac{\sqrt{537}}{8} x+\frac{5}{8}=-\frac{\sqrt{537}}{8}
Simplify.
x=\frac{\sqrt{537}-5}{8} x=\frac{-\sqrt{537}-5}{8}
Subtract \frac{5}{8} from both sides of the equation.