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4x^2+80x=150
We move all terms to the left:
4x^2+80x-(150)=0
a = 4; b = 80; c = -150;
Δ = b2-4ac
Δ = 802-4·4·(-150)
Δ = 8800
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{8800}=\sqrt{400*22}=\sqrt{400}*\sqrt{22}=20\sqrt{22}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(80)-20\sqrt{22}}{2*4}=\frac{-80-20\sqrt{22}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(80)+20\sqrt{22}}{2*4}=\frac{-80+20\sqrt{22}}{8} $
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