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20x(x+3)=625+10x
We move all terms to the left:
20x(x+3)-(625+10x)=0
We add all the numbers together, and all the variables
20x(x+3)-(10x+625)=0
We multiply parentheses
20x^2+60x-(10x+625)=0
We get rid of parentheses
20x^2+60x-10x-625=0
We add all the numbers together, and all the variables
20x^2+50x-625=0
a = 20; b = 50; c = -625;
Δ = b2-4ac
Δ = 502-4·20·(-625)
Δ = 52500
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{52500}=\sqrt{2500*21}=\sqrt{2500}*\sqrt{21}=50\sqrt{21}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(50)-50\sqrt{21}}{2*20}=\frac{-50-50\sqrt{21}}{40} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(50)+50\sqrt{21}}{2*20}=\frac{-50+50\sqrt{21}}{40} $
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