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4x^{2}+4x=5
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}+4x-5=5-5
Subtract 5 from both sides of the equation.
4x^{2}+4x-5=0
Subtracting 5 from itself leaves 0.
x=\frac{-4±\sqrt{4^{2}-4\times 4\left(-5\right)}}{2\times 4}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, 4 for b, and -5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-4±\sqrt{16-4\times 4\left(-5\right)}}{2\times 4}
Square 4.
x=\frac{-4±\sqrt{16-16\left(-5\right)}}{2\times 4}
Multiply -4 times 4.
x=\frac{-4±\sqrt{16+80}}{2\times 4}
Multiply -16 times -5.
x=\frac{-4±\sqrt{96}}{2\times 4}
Add 16 to 80.
x=\frac{-4±4\sqrt{6}}{2\times 4}
Take the square root of 96.
x=\frac{-4±4\sqrt{6}}{8}
Multiply 2 times 4.
x=\frac{4\sqrt{6}-4}{8}
Now solve the equation x=\frac{-4±4\sqrt{6}}{8} when ± is plus. Add -4 to 4\sqrt{6}.
x=\frac{\sqrt{6}-1}{2}
Divide -4+4\sqrt{6} by 8.
x=\frac{-4\sqrt{6}-4}{8}
Now solve the equation x=\frac{-4±4\sqrt{6}}{8} when ± is minus. Subtract 4\sqrt{6} from -4.
x=\frac{-\sqrt{6}-1}{2}
Divide -4-4\sqrt{6} by 8.
x=\frac{\sqrt{6}-1}{2} x=\frac{-\sqrt{6}-1}{2}
The equation is now solved.
4x^{2}+4x=5
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.
\frac{4x^{2}+4x}{4}=\frac{5}{4}
Divide both sides by 4.
x^{2}+\frac{4}{4}x=\frac{5}{4}
Dividing by 4 undoes the multiplication by 4.
x^{2}+x=\frac{5}{4}
Divide 4 by 4.
x^{2}+x+\left(\frac{1}{2}\right)^{2}=\frac{5}{4}+\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{5+1}{4}
Square \frac{1}{2} by squaring both the numerator and the denominator of the fraction.
x^{2}+x+\frac{1}{4}=\frac{3}{2}
Add \frac{5}{4} 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{3}{2}
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{3}{2}}
Take the square root of both sides of the equation.
x+\frac{1}{2}=\frac{\sqrt{6}}{2} x+\frac{1}{2}=-\frac{\sqrt{6}}{2}
Simplify.
x=\frac{\sqrt{6}-1}{2} x=\frac{-\sqrt{6}-1}{2}
Subtract \frac{1}{2} from both sides of the equation.