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10x^2+40x-600=0
a = 10; b = 40; c = -600;
Δ = b2-4ac
Δ = 402-4·10·(-600)
Δ = 25600
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{25600}=160$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-160}{2*10}=\frac{-200}{20} =-10 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+160}{2*10}=\frac{120}{20} =6 $
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