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use quadratic formula to solve. 2x^2-3x=-4
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To use the quadratic you get it equal to 0 then you have your a, b, and c coefficients to plug in.
Quadratic equation: \[ax^2 + bx +c = 0\rightarrow x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\]
\[2x^2-3x=-4 \rightarrow 2x^2-3x+4=0 \rightarrow a=2, b=-3, c=4 \] \[\rightarrow x = {{-b \pm \sqrt{b^2-4 \cdot a \cdot c}}\over{2 \cdot a}} = {{-(-3) \pm \sqrt{(-3)^2-4 \cdot 2 \cdot 4}}\over{2 \cdot 2}} \] \[= {{3 \pm \sqrt{9-32}}\over{4}} = {{3 \pm \sqrt{-23}}\over{4}} = {3 \over 4}\pm{\sqrt{23}\over{4}}\cdot i\] See also http://wolframalpha.com/input/?i=2x^2-3x%3D-4
U can also use x1+x2=-b/a and x1*x2=c/a, Viete, assuming you speak Slovak.
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