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3x^{2}+2x-3=0
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=\frac{-2±\sqrt{2^{2}-4\times 3\left(-3\right)}}{2\times 3}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 2 for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-2±\sqrt{4-4\times 3\left(-3\right)}}{2\times 3}
Square 2.
x=\frac{-2±\sqrt{4-12\left(-3\right)}}{2\times 3}
Multiply -4 times 3.
x=\frac{-2±\sqrt{4+36}}{2\times 3}
Multiply -12 times -3.
x=\frac{-2±\sqrt{40}}{2\times 3}
Add 4 to 36.
x=\frac{-2±2\sqrt{10}}{2\times 3}
Take the square root of 40.
x=\frac{-2±2\sqrt{10}}{6}
Multiply 2 times 3.
x=\frac{2\sqrt{10}-2}{6}
Now solve the equation x=\frac{-2±2\sqrt{10}}{6} when ± is plus. Add -2 to 2\sqrt{10}.
x=\frac{\sqrt{10}-1}{3}
Divide -2+2\sqrt{10} by 6.
x=\frac{-2\sqrt{10}-2}{6}
Now solve the equation x=\frac{-2±2\sqrt{10}}{6} when ± is minus. Subtract 2\sqrt{10} from -2.
x=\frac{-\sqrt{10}-1}{3}
Divide -2-2\sqrt{10} by 6.
x=\frac{\sqrt{10}-1}{3} x=\frac{-\sqrt{10}-1}{3}
The equation is now solved.
3x^{2}+2x-3=0
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}+2x-3-\left(-3\right)=-\left(-3\right)
Add 3 to both sides of the equation.
3x^{2}+2x=-\left(-3\right)
Subtracting -3 from itself leaves 0.
3x^{2}+2x=3
Subtract -3 from 0.
\frac{3x^{2}+2x}{3}=\frac{3}{3}
Divide both sides by 3.
x^{2}+\frac{2}{3}x=\frac{3}{3}
Dividing by 3 undoes the multiplication by 3.
x^{2}+\frac{2}{3}x=1
Divide 3 by 3.
x^{2}+\frac{2}{3}x+\left(\frac{1}{3}\right)^{2}=1+\left(\frac{1}{3}\right)^{2}
Divide \frac{2}{3}, the coefficient of the x term, by 2 to get \frac{1}{3}. Then add the square of \frac{1}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+\frac{2}{3}x+\frac{1}{9}=1+\frac{1}{9}
Square \frac{1}{3} by squaring both the numerator and the denominator of the fraction.
x^{2}+\frac{2}{3}x+\frac{1}{9}=\frac{10}{9}
Add 1 to \frac{1}{9}.
\left(x+\frac{1}{3}\right)^{2}=\frac{10}{9}
Factor x^{2}+\frac{2}{3}x+\frac{1}{9}. 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}{3}\right)^{2}}=\sqrt{\frac{10}{9}}
Take the square root of both sides of the equation.
x+\frac{1}{3}=\frac{\sqrt{10}}{3} x+\frac{1}{3}=-\frac{\sqrt{10}}{3}
Simplify.
x=\frac{\sqrt{10}-1}{3} x=\frac{-\sqrt{10}-1}{3}
Subtract \frac{1}{3} from both sides of the equation.