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5x^2+41x+8=0
a = 5; b = 41; c = +8;
Δ = b2-4ac
Δ = 412-4·5·8
Δ = 1521
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{1521}=39$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(41)-39}{2*5}=\frac{-80}{10} =-8 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(41)+39}{2*5}=\frac{-2}{10} =-1/5 $
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