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2x^2+45x-40=0
a = 2; b = 45; c = -40;
Δ = b2-4ac
Δ = 452-4·2·(-40)
Δ = 2345
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{-(45)-\sqrt{2345}}{2*2}=\frac{-45-\sqrt{2345}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(45)+\sqrt{2345}}{2*2}=\frac{-45+\sqrt{2345}}{4} $
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