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2x^2-8x-42=0
a = 2; b = -8; c = -42;
Δ = b2-4ac
Δ = -82-4·2·(-42)
Δ = 400
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{400}=20$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-20}{2*2}=\frac{-12}{4} =-3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+20}{2*2}=\frac{28}{4} =7 $
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