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x2+40x-176=0
We add all the numbers together, and all the variables
x^2+40x-176=0
a = 1; b = 40; c = -176;
Δ = b2-4ac
Δ = 402-4·1·(-176)
Δ = 2304
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{2304}=48$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-48}{2*1}=\frac{-88}{2} =-44 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+48}{2*1}=\frac{8}{2} =4 $
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