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z^{2}-\frac{3}{2}z+\frac{9}{16}=0
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.
z=\frac{-\left(-\frac{3}{2}\right)±\sqrt{\left(-\frac{3}{2}\right)^{2}-4\times \frac{9}{16}}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -\frac{3}{2} for b, and \frac{9}{16} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
z=\frac{-\left(-\frac{3}{2}\right)±\sqrt{\frac{9}{4}-4\times \frac{9}{16}}}{2}
Square -\frac{3}{2} by squaring both the numerator and the denominator of the fraction.
z=\frac{-\left(-\frac{3}{2}\right)±\sqrt{\frac{9-9}{4}}}{2}
Multiply -4 times \frac{9}{16}.
z=\frac{-\left(-\frac{3}{2}\right)±\sqrt{0}}{2}
Add \frac{9}{4} to -\frac{9}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
z=-\frac{-\frac{3}{2}}{2}
Take the square root of 0.
z=\frac{\frac{3}{2}}{2}
The opposite of -\frac{3}{2} is \frac{3}{2}.
z=\frac{3}{4}
Divide \frac{3}{2} by 2.
z^{2}-\frac{3}{2}z+\frac{9}{16}=0
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.
\left(z-\frac{3}{4}\right)^{2}=0
Factor z^{2}-\frac{3}{2}z+\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(z-\frac{3}{4}\right)^{2}}=\sqrt{0}
Take the square root of both sides of the equation.
z-\frac{3}{4}=0 z-\frac{3}{4}=0
Simplify.
z=\frac{3}{4} z=\frac{3}{4}
Add \frac{3}{4} to both sides of the equation.
z=\frac{3}{4}
The equation is now solved. Solutions are the same.