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Solve for h
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Solve for x (complex solution)
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2h^{2}+3\left(x+h-x\right)=5
Combine x and -x to get 0.
2h^{2}+3h=5
Combine x and -x to get 0.
2h^{2}+3h-5=0
Subtract 5 from both sides.
h=\frac{-3±\sqrt{3^{2}-4\times 2\left(-5\right)}}{2\times 2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 3 for b, and -5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
h=\frac{-3±\sqrt{9-4\times 2\left(-5\right)}}{2\times 2}
Square 3.
h=\frac{-3±\sqrt{9-8\left(-5\right)}}{2\times 2}
Multiply -4 times 2.
h=\frac{-3±\sqrt{9+40}}{2\times 2}
Multiply -8 times -5.
h=\frac{-3±\sqrt{49}}{2\times 2}
Add 9 to 40.
h=\frac{-3±7}{2\times 2}
Take the square root of 49.
h=\frac{-3±7}{4}
Multiply 2 times 2.
h=\frac{4}{4}
Now solve the equation h=\frac{-3±7}{4} when ± is plus. Add -3 to 7.
h=1
Divide 4 by 4.
h=-\frac{10}{4}
Now solve the equation h=\frac{-3±7}{4} when ± is minus. Subtract 7 from -3.
h=-\frac{5}{2}
Reduce the fraction \frac{-10}{4} to lowest terms by extracting and canceling out 2.
h=1 h=-\frac{5}{2}
The equation is now solved.
2h^{2}+3\left(x+h-x\right)=5
Combine x and -x to get 0.
2h^{2}+3h=5
Combine x and -x to get 0.
\frac{2h^{2}+3h}{2}=\frac{5}{2}
Divide both sides by 2.
h^{2}+\frac{3}{2}h=\frac{5}{2}
Dividing by 2 undoes the multiplication by 2.
h^{2}+\frac{3}{2}h+\left(\frac{3}{4}\right)^{2}=\frac{5}{2}+\left(\frac{3}{4}\right)^{2}
Divide \frac{3}{2}, the coefficient of the x term, by 2 to get \frac{3}{4}. Then add the square of \frac{3}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
h^{2}+\frac{3}{2}h+\frac{9}{16}=\frac{5}{2}+\frac{9}{16}
Square \frac{3}{4} by squaring both the numerator and the denominator of the fraction.
h^{2}+\frac{3}{2}h+\frac{9}{16}=\frac{49}{16}
Add \frac{5}{2} to \frac{9}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(h+\frac{3}{4}\right)^{2}=\frac{49}{16}
Factor h^{2}+\frac{3}{2}h+\frac{9}{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(h+\frac{3}{4}\right)^{2}}=\sqrt{\frac{49}{16}}
Take the square root of both sides of the equation.
h+\frac{3}{4}=\frac{7}{4} h+\frac{3}{4}=-\frac{7}{4}
Simplify.
h=1 h=-\frac{5}{2}
Subtract \frac{3}{4} from both sides of the equation.