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-4z^2+13z-3=0
a = -4; b = 13; c = -3;
Δ = b2-4ac
Δ = 132-4·(-4)·(-3)
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{121}=11$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-11}{2*-4}=\frac{-24}{-8} =+3 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+11}{2*-4}=\frac{-2}{-8} =1/4 $
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