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3x^{2}-3x+4=12
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}-3x+4-12=12-12
Subtract 12 from both sides of the equation.
3x^{2}-3x+4-12=0
Subtracting 12 from itself leaves 0.
3x^{2}-3x-8=0
Subtract 12 from 4.
x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 3\left(-8\right)}}{2\times 3}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -3 for b, and -8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-3\right)±\sqrt{9-4\times 3\left(-8\right)}}{2\times 3}
Square -3.
x=\frac{-\left(-3\right)±\sqrt{9-12\left(-8\right)}}{2\times 3}
Multiply -4 times 3.
x=\frac{-\left(-3\right)±\sqrt{9+96}}{2\times 3}
Multiply -12 times -8.
x=\frac{-\left(-3\right)±\sqrt{105}}{2\times 3}
Add 9 to 96.
x=\frac{3±\sqrt{105}}{2\times 3}
The opposite of -3 is 3.
x=\frac{3±\sqrt{105}}{6}
Multiply 2 times 3.
x=\frac{\sqrt{105}+3}{6}
Now solve the equation x=\frac{3±\sqrt{105}}{6} when ± is plus. Add 3 to \sqrt{105}.
x=\frac{\sqrt{105}}{6}+\frac{1}{2}
Divide 3+\sqrt{105} by 6.
x=\frac{3-\sqrt{105}}{6}
Now solve the equation x=\frac{3±\sqrt{105}}{6} when ± is minus. Subtract \sqrt{105} from 3.
x=-\frac{\sqrt{105}}{6}+\frac{1}{2}
Divide 3-\sqrt{105} by 6.
x=\frac{\sqrt{105}}{6}+\frac{1}{2} x=-\frac{\sqrt{105}}{6}+\frac{1}{2}
The equation is now solved.
3x^{2}-3x+4=12
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}-3x+4-4=12-4
Subtract 4 from both sides of the equation.
3x^{2}-3x=12-4
Subtracting 4 from itself leaves 0.
3x^{2}-3x=8
Subtract 4 from 12.
\frac{3x^{2}-3x}{3}=\frac{8}{3}
Divide both sides by 3.
x^{2}+\left(-\frac{3}{3}\right)x=\frac{8}{3}
Dividing by 3 undoes the multiplication by 3.
x^{2}-x=\frac{8}{3}
Divide -3 by 3.
x^{2}-x+\left(-\frac{1}{2}\right)^{2}=\frac{8}{3}+\left(-\frac{1}{2}\right)^{2}
Divide -1, the coefficient of the x term, by 2 to get -\frac{1}{2}. Then add the square of -\frac{1}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-x+\frac{1}{4}=\frac{8}{3}+\frac{1}{4}
Square -\frac{1}{2} by squaring both the numerator and the denominator of the fraction.
x^{2}-x+\frac{1}{4}=\frac{35}{12}
Add \frac{8}{3} to \frac{1}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x-\frac{1}{2}\right)^{2}=\frac{35}{12}
Factor x^{2}-x+\frac{1}{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{1}{2}\right)^{2}}=\sqrt{\frac{35}{12}}
Take the square root of both sides of the equation.
x-\frac{1}{2}=\frac{\sqrt{105}}{6} x-\frac{1}{2}=-\frac{\sqrt{105}}{6}
Simplify.
x=\frac{\sqrt{105}}{6}+\frac{1}{2} x=-\frac{\sqrt{105}}{6}+\frac{1}{2}
Add \frac{1}{2} to both sides of the equation.