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\frac{5}{2}x^{2}+3x=1.2
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.
\frac{5}{2}x^{2}+3x-1.2=1.2-1.2
Subtract 1.2 from both sides of the equation.
\frac{5}{2}x^{2}+3x-1.2=0
Subtracting 1.2 from itself leaves 0.
x=\frac{-3±\sqrt{3^{2}-4\times \frac{5}{2}\left(-1.2\right)}}{2\times \frac{5}{2}}
This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{5}{2} for a, 3 for b, and -1.2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-3±\sqrt{9-4\times \frac{5}{2}\left(-1.2\right)}}{2\times \frac{5}{2}}
Square 3.
x=\frac{-3±\sqrt{9-10\left(-1.2\right)}}{2\times \frac{5}{2}}
Multiply -4 times \frac{5}{2}.
x=\frac{-3±\sqrt{9+12}}{2\times \frac{5}{2}}
Multiply -10 times -1.2.
x=\frac{-3±\sqrt{21}}{2\times \frac{5}{2}}
Add 9 to 12.
x=\frac{-3±\sqrt{21}}{5}
Multiply 2 times \frac{5}{2}.
x=\frac{\sqrt{21}-3}{5}
Now solve the equation x=\frac{-3±\sqrt{21}}{5} when ± is plus. Add -3 to \sqrt{21}.
x=\frac{-\sqrt{21}-3}{5}
Now solve the equation x=\frac{-3±\sqrt{21}}{5} when ± is minus. Subtract \sqrt{21} from -3.
x=\frac{\sqrt{21}-3}{5} x=\frac{-\sqrt{21}-3}{5}
The equation is now solved.
\frac{5}{2}x^{2}+3x=1.2
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{\frac{5}{2}x^{2}+3x}{\frac{5}{2}}=\frac{1.2}{\frac{5}{2}}
Divide both sides of the equation by \frac{5}{2}, which is the same as multiplying both sides by the reciprocal of the fraction.
x^{2}+\frac{3}{\frac{5}{2}}x=\frac{1.2}{\frac{5}{2}}
Dividing by \frac{5}{2} undoes the multiplication by \frac{5}{2}.
x^{2}+\frac{6}{5}x=\frac{1.2}{\frac{5}{2}}
Divide 3 by \frac{5}{2} by multiplying 3 by the reciprocal of \frac{5}{2}.
x^{2}+\frac{6}{5}x=\frac{12}{25}
Divide 1.2 by \frac{5}{2} by multiplying 1.2 by the reciprocal of \frac{5}{2}.
x^{2}+\frac{6}{5}x+\left(\frac{3}{5}\right)^{2}=\frac{12}{25}+\left(\frac{3}{5}\right)^{2}
Divide \frac{6}{5}, the coefficient of the x term, by 2 to get \frac{3}{5}. Then add the square of \frac{3}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+\frac{6}{5}x+\frac{9}{25}=\frac{12+9}{25}
Square \frac{3}{5} by squaring both the numerator and the denominator of the fraction.
x^{2}+\frac{6}{5}x+\frac{9}{25}=\frac{21}{25}
Add \frac{12}{25} to \frac{9}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x+\frac{3}{5}\right)^{2}=\frac{21}{25}
Factor x^{2}+\frac{6}{5}x+\frac{9}{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{3}{5}\right)^{2}}=\sqrt{\frac{21}{25}}
Take the square root of both sides of the equation.
x+\frac{3}{5}=\frac{\sqrt{21}}{5} x+\frac{3}{5}=-\frac{\sqrt{21}}{5}
Simplify.
x=\frac{\sqrt{21}-3}{5} x=\frac{-\sqrt{21}-3}{5}
Subtract \frac{3}{5} from both sides of the equation.