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5x^2+32x+64=2704
We move all terms to the left:
5x^2+32x+64-(2704)=0
We add all the numbers together, and all the variables
5x^2+32x-2640=0
a = 5; b = 32; c = -2640;
Δ = b2-4ac
Δ = 322-4·5·(-2640)
Δ = 53824
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{53824}=232$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-232}{2*5}=\frac{-264}{10} =-26+2/5 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+232}{2*5}=\frac{200}{10} =20 $
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