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40x^2-38x-15=0
a = 40; b = -38; c = -15;
Δ = b2-4ac
Δ = -382-4·40·(-15)
Δ = 3844
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{3844}=62$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-38)-62}{2*40}=\frac{-24}{80} =-3/10 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-38)+62}{2*40}=\frac{100}{80} =1+1/4 $
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