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(5x-3)(4x+1)=0
We multiply parentheses ..
(+20x^2+5x-12x-3)=0
We get rid of parentheses
20x^2+5x-12x-3=0
We add all the numbers together, and all the variables
20x^2-7x-3=0
a = 20; b = -7; c = -3;
Δ = b2-4ac
Δ = -72-4·20·(-3)
Δ = 289
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{289}=17$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-17}{2*20}=\frac{-10}{40} =-1/4 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+17}{2*20}=\frac{24}{40} =3/5 $
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