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8x^2+41x-42=0
a = 8; b = 41; c = -42;
Δ = b2-4ac
Δ = 412-4·8·(-42)
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(41)-55}{2*8}=\frac{-96}{16} =-6 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(41)+55}{2*8}=\frac{14}{16} =7/8 $
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