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27x+x^2-160=0
a = 1; b = 27; c = -160;
Δ = b2-4ac
Δ = 272-4·1·(-160)
Δ = 1369
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{1369}=37$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(27)-37}{2*1}=\frac{-64}{2} =-32 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(27)+37}{2*1}=\frac{10}{2} =5 $
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