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x^2+42x-190=0
a = 1; b = 42; c = -190;
Δ = b2-4ac
Δ = 422-4·1·(-190)
Δ = 2524
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2524}=\sqrt{4*631}=\sqrt{4}*\sqrt{631}=2\sqrt{631}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(42)-2\sqrt{631}}{2*1}=\frac{-42-2\sqrt{631}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(42)+2\sqrt{631}}{2*1}=\frac{-42+2\sqrt{631}}{2} $
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