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5x(4x+3)=20x+15x
We move all terms to the left:
5x(4x+3)-(20x+15x)=0
We add all the numbers together, and all the variables
5x(4x+3)-(+35x)=0
We multiply parentheses
20x^2+15x-(+35x)=0
We get rid of parentheses
20x^2+15x-35x=0
We add all the numbers together, and all the variables
20x^2-20x=0
a = 20; b = -20; c = 0;
Δ = b2-4ac
Δ = -202-4·20·0
Δ = 400
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{400}=20$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-20}{2*20}=\frac{0}{40} =0 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+20}{2*20}=\frac{40}{40} =1 $
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