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5x^{2}+2x+3=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.
5x^{2}+2x+3-7=7-7
Subtract 7 from both sides of the equation.
5x^{2}+2x+3-7=0
Subtracting 7 from itself leaves 0.
5x^{2}+2x-4=0
Subtract 7 from 3.
x=\frac{-2±\sqrt{2^{2}-4\times 5\left(-4\right)}}{2\times 5}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, 2 for b, and -4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-2±\sqrt{4-4\times 5\left(-4\right)}}{2\times 5}
Square 2.
x=\frac{-2±\sqrt{4-20\left(-4\right)}}{2\times 5}
Multiply -4 times 5.
x=\frac{-2±\sqrt{4+80}}{2\times 5}
Multiply -20 times -4.
x=\frac{-2±\sqrt{84}}{2\times 5}
Add 4 to 80.
x=\frac{-2±2\sqrt{21}}{2\times 5}
Take the square root of 84.
x=\frac{-2±2\sqrt{21}}{10}
Multiply 2 times 5.
x=\frac{2\sqrt{21}-2}{10}
Now solve the equation x=\frac{-2±2\sqrt{21}}{10} when ± is plus. Add -2 to 2\sqrt{21}.
x=\frac{\sqrt{21}-1}{5}
Divide -2+2\sqrt{21} by 10.
x=\frac{-2\sqrt{21}-2}{10}
Now solve the equation x=\frac{-2±2\sqrt{21}}{10} when ± is minus. Subtract 2\sqrt{21} from -2.
x=\frac{-\sqrt{21}-1}{5}
Divide -2-2\sqrt{21} by 10.
x=\frac{\sqrt{21}-1}{5} x=\frac{-\sqrt{21}-1}{5}
The equation is now solved.
5x^{2}+2x+3=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.
5x^{2}+2x+3-3=7-3
Subtract 3 from both sides of the equation.
5x^{2}+2x=7-3
Subtracting 3 from itself leaves 0.
5x^{2}+2x=4
Subtract 3 from 7.
\frac{5x^{2}+2x}{5}=\frac{4}{5}
Divide both sides by 5.
x^{2}+\frac{2}{5}x=\frac{4}{5}
Dividing by 5 undoes the multiplication by 5.
x^{2}+\frac{2}{5}x+\left(\frac{1}{5}\right)^{2}=\frac{4}{5}+\left(\frac{1}{5}\right)^{2}
Divide \frac{2}{5}, the coefficient of the x term, by 2 to get \frac{1}{5}. Then add the square of \frac{1}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+\frac{2}{5}x+\frac{1}{25}=\frac{4}{5}+\frac{1}{25}
Square \frac{1}{5} by squaring both the numerator and the denominator of the fraction.
x^{2}+\frac{2}{5}x+\frac{1}{25}=\frac{21}{25}
Add \frac{4}{5} to \frac{1}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x+\frac{1}{5}\right)^{2}=\frac{21}{25}
Factor x^{2}+\frac{2}{5}x+\frac{1}{25}. 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}{5}\right)^{2}}=\sqrt{\frac{21}{25}}
Take the square root of both sides of the equation.
x+\frac{1}{5}=\frac{\sqrt{21}}{5} x+\frac{1}{5}=-\frac{\sqrt{21}}{5}
Simplify.
x=\frac{\sqrt{21}-1}{5} x=\frac{-\sqrt{21}-1}{5}
Subtract \frac{1}{5} from both sides of the equation.