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4x^2+40x-224=0
a = 4; b = 40; c = -224;
Δ = b2-4ac
Δ = 402-4·4·(-224)
Δ = 5184
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{5184}=72$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-72}{2*4}=\frac{-112}{8} =-14 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+72}{2*4}=\frac{32}{8} =4 $
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