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4x(5x+4)=5x+4
We move all terms to the left:
4x(5x+4)-(5x+4)=0
We multiply parentheses
20x^2+16x-(5x+4)=0
We get rid of parentheses
20x^2+16x-5x-4=0
We add all the numbers together, and all the variables
20x^2+11x-4=0
a = 20; b = 11; c = -4;
Δ = b2-4ac
Δ = 112-4·20·(-4)
Δ = 441
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{441}=21$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-21}{2*20}=\frac{-32}{40} =-4/5 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+21}{2*20}=\frac{10}{40} =1/4 $
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