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x^2+42x-135=0
a = 1; b = 42; c = -135;
Δ = b2-4ac
Δ = 422-4·1·(-135)
Δ = 2304
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{2304}=48$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(42)-48}{2*1}=\frac{-90}{2} =-45 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(42)+48}{2*1}=\frac{6}{2} =3 $
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