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x^{2}+3x-16=48
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^{2}+3x-16-48=48-48
Subtract 48 from both sides of the equation.
x^{2}+3x-16-48=0
Subtracting 48 from itself leaves 0.
x^{2}+3x-64=0
Subtract 48 from -16.
x=\frac{-3±\sqrt{3^{2}-4\left(-64\right)}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 3 for b, and -64 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-3±\sqrt{9-4\left(-64\right)}}{2}
Square 3.
x=\frac{-3±\sqrt{9+256}}{2}
Multiply -4 times -64.
x=\frac{-3±\sqrt{265}}{2}
Add 9 to 256.
x=\frac{\sqrt{265}-3}{2}
Now solve the equation x=\frac{-3±\sqrt{265}}{2} when ± is plus. Add -3 to \sqrt{265}.
x=\frac{-\sqrt{265}-3}{2}
Now solve the equation x=\frac{-3±\sqrt{265}}{2} when ± is minus. Subtract \sqrt{265} from -3.
x=\frac{\sqrt{265}-3}{2} x=\frac{-\sqrt{265}-3}{2}
The equation is now solved.
x^{2}+3x-16=48
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}+3x-16-\left(-16\right)=48-\left(-16\right)
Add 16 to both sides of the equation.
x^{2}+3x=48-\left(-16\right)
Subtracting -16 from itself leaves 0.
x^{2}+3x=64
Subtract -16 from 48.
x^{2}+3x+\left(\frac{3}{2}\right)^{2}=64+\left(\frac{3}{2}\right)^{2}
Divide 3, the coefficient of the x term, by 2 to get \frac{3}{2}. Then add the square of \frac{3}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+3x+\frac{9}{4}=64+\frac{9}{4}
Square \frac{3}{2} by squaring both the numerator and the denominator of the fraction.
x^{2}+3x+\frac{9}{4}=\frac{265}{4}
Add 64 to \frac{9}{4}.
\left(x+\frac{3}{2}\right)^{2}=\frac{265}{4}
Factor x^{2}+3x+\frac{9}{4}. 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}{2}\right)^{2}}=\sqrt{\frac{265}{4}}
Take the square root of both sides of the equation.
x+\frac{3}{2}=\frac{\sqrt{265}}{2} x+\frac{3}{2}=-\frac{\sqrt{265}}{2}
Simplify.
x=\frac{\sqrt{265}-3}{2} x=\frac{-\sqrt{265}-3}{2}
Subtract \frac{3}{2} from both sides of the equation.