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43.57(x+20)x3=305
We move all terms to the left:
43.57(x+20)x3-(305)=0
We multiply parentheses
43x^2+860x-305=0
a = 43; b = 860; c = -305;
Δ = b2-4ac
Δ = 8602-4·43·(-305)
Δ = 792060
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{792060}=\sqrt{4*198015}=\sqrt{4}*\sqrt{198015}=2\sqrt{198015}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(860)-2\sqrt{198015}}{2*43}=\frac{-860-2\sqrt{198015}}{86} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(860)+2\sqrt{198015}}{2*43}=\frac{-860+2\sqrt{198015}}{86} $
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