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10x^2+42x+8=0
a = 10; b = 42; c = +8;
Δ = b2-4ac
Δ = 422-4·10·8
Δ = 1444
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{1444}=38$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(42)-38}{2*10}=\frac{-80}{20} =-4 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(42)+38}{2*10}=\frac{-4}{20} =-1/5 $
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