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-3x^2+25x-36=0
a = -3; b = 25; c = -36;
Δ = b2-4ac
Δ = 252-4·(-3)·(-36)
Δ = 193
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-\sqrt{193}}{2*-3}=\frac{-25-\sqrt{193}}{-6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+\sqrt{193}}{2*-3}=\frac{-25+\sqrt{193}}{-6} $
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