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20x^2+15x-35=0
a = 20; b = 15; c = -35;
Δ = b2-4ac
Δ = 152-4·20·(-35)
Δ = 3025
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{3025}=55$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-55}{2*20}=\frac{-70}{40} =-1+3/4 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+55}{2*20}=\frac{40}{40} =1 $
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