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4x^{2}+2x+3=8
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}+2x+3-8=8-8
Subtract 8 from both sides of the equation.
4x^{2}+2x+3-8=0
Subtracting 8 from itself leaves 0.
4x^{2}+2x-5=0
Subtract 8 from 3.
x=\frac{-2±\sqrt{2^{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, 2 for b, and -5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-2±\sqrt{4-4\times 4\left(-5\right)}}{2\times 4}
Square 2.
x=\frac{-2±\sqrt{4-16\left(-5\right)}}{2\times 4}
Multiply -4 times 4.
x=\frac{-2±\sqrt{4+80}}{2\times 4}
Multiply -16 times -5.
x=\frac{-2±\sqrt{84}}{2\times 4}
Add 4 to 80.
x=\frac{-2±2\sqrt{21}}{2\times 4}
Take the square root of 84.
x=\frac{-2±2\sqrt{21}}{8}
Multiply 2 times 4.
x=\frac{2\sqrt{21}-2}{8}
Now solve the equation x=\frac{-2±2\sqrt{21}}{8} when ± is plus. Add -2 to 2\sqrt{21}.
x=\frac{\sqrt{21}-1}{4}
Divide -2+2\sqrt{21} by 8.
x=\frac{-2\sqrt{21}-2}{8}
Now solve the equation x=\frac{-2±2\sqrt{21}}{8} when ± is minus. Subtract 2\sqrt{21} from -2.
x=\frac{-\sqrt{21}-1}{4}
Divide -2-2\sqrt{21} by 8.
x=\frac{\sqrt{21}-1}{4} x=\frac{-\sqrt{21}-1}{4}
The equation is now solved.
4x^{2}+2x+3=8
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.
4x^{2}+2x+3-3=8-3
Subtract 3 from both sides of the equation.
4x^{2}+2x=8-3
Subtracting 3 from itself leaves 0.
4x^{2}+2x=5
Subtract 3 from 8.
\frac{4x^{2}+2x}{4}=\frac{5}{4}
Divide both sides by 4.
x^{2}+\frac{2}{4}x=\frac{5}{4}
Dividing by 4 undoes the multiplication by 4.
x^{2}+\frac{1}{2}x=\frac{5}{4}
Reduce the fraction \frac{2}{4} to lowest terms by extracting and canceling out 2.
x^{2}+\frac{1}{2}x+\left(\frac{1}{4}\right)^{2}=\frac{5}{4}+\left(\frac{1}{4}\right)^{2}
Divide \frac{1}{2}, the coefficient of the x term, by 2 to get \frac{1}{4}. Then add the square of \frac{1}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+\frac{1}{2}x+\frac{1}{16}=\frac{5}{4}+\frac{1}{16}
Square \frac{1}{4} by squaring both the numerator and the denominator of the fraction.
x^{2}+\frac{1}{2}x+\frac{1}{16}=\frac{21}{16}
Add \frac{5}{4} to \frac{1}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x+\frac{1}{4}\right)^{2}=\frac{21}{16}
Factor x^{2}+\frac{1}{2}x+\frac{1}{16}. 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}{4}\right)^{2}}=\sqrt{\frac{21}{16}}
Take the square root of both sides of the equation.
x+\frac{1}{4}=\frac{\sqrt{21}}{4} x+\frac{1}{4}=-\frac{\sqrt{21}}{4}
Simplify.
x=\frac{\sqrt{21}-1}{4} x=\frac{-\sqrt{21}-1}{4}
Subtract \frac{1}{4} from both sides of the equation.