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3x^{2}+\frac{3}{4}x+9=7
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.
3x^{2}+\frac{3}{4}x+9-7=7-7
Subtract 7 from both sides of the equation.
3x^{2}+\frac{3}{4}x+9-7=0
Subtracting 7 from itself leaves 0.
3x^{2}+\frac{3}{4}x+2=0
Subtract 7 from 9.
x=\frac{-\frac{3}{4}±\sqrt{\left(\frac{3}{4}\right)^{2}-4\times 3\times 2}}{2\times 3}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, \frac{3}{4} for b, and 2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\frac{3}{4}±\sqrt{\frac{9}{16}-4\times 3\times 2}}{2\times 3}
Square \frac{3}{4} by squaring both the numerator and the denominator of the fraction.
x=\frac{-\frac{3}{4}±\sqrt{\frac{9}{16}-12\times 2}}{2\times 3}
Multiply -4 times 3.
x=\frac{-\frac{3}{4}±\sqrt{\frac{9}{16}-24}}{2\times 3}
Multiply -12 times 2.
x=\frac{-\frac{3}{4}±\sqrt{-\frac{375}{16}}}{2\times 3}
Add \frac{9}{16} to -24.
x=\frac{-\frac{3}{4}±\frac{5\sqrt{15}i}{4}}{2\times 3}
Take the square root of -\frac{375}{16}.
x=\frac{-\frac{3}{4}±\frac{5\sqrt{15}i}{4}}{6}
Multiply 2 times 3.
x=\frac{-3+5\sqrt{15}i}{4\times 6}
Now solve the equation x=\frac{-\frac{3}{4}±\frac{5\sqrt{15}i}{4}}{6} when ± is plus. Add -\frac{3}{4} to \frac{5i\sqrt{15}}{4}.
x=\frac{5\sqrt{15}i}{24}-\frac{1}{8}
Divide \frac{-3+5i\sqrt{15}}{4} by 6.
x=\frac{-5\sqrt{15}i-3}{4\times 6}
Now solve the equation x=\frac{-\frac{3}{4}±\frac{5\sqrt{15}i}{4}}{6} when ± is minus. Subtract \frac{5i\sqrt{15}}{4} from -\frac{3}{4}.
x=-\frac{5\sqrt{15}i}{24}-\frac{1}{8}
Divide \frac{-3-5i\sqrt{15}}{4} by 6.
x=\frac{5\sqrt{15}i}{24}-\frac{1}{8} x=-\frac{5\sqrt{15}i}{24}-\frac{1}{8}
The equation is now solved.
3x^{2}+\frac{3}{4}x+9=7
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.
3x^{2}+\frac{3}{4}x+9-9=7-9
Subtract 9 from both sides of the equation.
3x^{2}+\frac{3}{4}x=7-9
Subtracting 9 from itself leaves 0.
3x^{2}+\frac{3}{4}x=-2
Subtract 9 from 7.
\frac{3x^{2}+\frac{3}{4}x}{3}=-\frac{2}{3}
Divide both sides by 3.
x^{2}+\frac{\frac{3}{4}}{3}x=-\frac{2}{3}
Dividing by 3 undoes the multiplication by 3.
x^{2}+\frac{1}{4}x=-\frac{2}{3}
Divide \frac{3}{4} by 3.
x^{2}+\frac{1}{4}x+\left(\frac{1}{8}\right)^{2}=-\frac{2}{3}+\left(\frac{1}{8}\right)^{2}
Divide \frac{1}{4}, the coefficient of the x term, by 2 to get \frac{1}{8}. Then add the square of \frac{1}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+\frac{1}{4}x+\frac{1}{64}=-\frac{2}{3}+\frac{1}{64}
Square \frac{1}{8} by squaring both the numerator and the denominator of the fraction.
x^{2}+\frac{1}{4}x+\frac{1}{64}=-\frac{125}{192}
Add -\frac{2}{3} to \frac{1}{64} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x+\frac{1}{8}\right)^{2}=-\frac{125}{192}
Factor x^{2}+\frac{1}{4}x+\frac{1}{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{1}{8}\right)^{2}}=\sqrt{-\frac{125}{192}}
Take the square root of both sides of the equation.
x+\frac{1}{8}=\frac{5\sqrt{15}i}{24} x+\frac{1}{8}=-\frac{5\sqrt{15}i}{24}
Simplify.
x=\frac{5\sqrt{15}i}{24}-\frac{1}{8} x=-\frac{5\sqrt{15}i}{24}-\frac{1}{8}
Subtract \frac{1}{8} from both sides of the equation.