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(5x-4)(3x+12)=0
We multiply parentheses ..
(+15x^2+60x-12x-48)=0
We get rid of parentheses
15x^2+60x-12x-48=0
We add all the numbers together, and all the variables
15x^2+48x-48=0
a = 15; b = 48; c = -48;
Δ = b2-4ac
Δ = 482-4·15·(-48)
Δ = 5184
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{5184}=72$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(48)-72}{2*15}=\frac{-120}{30} =-4 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(48)+72}{2*15}=\frac{24}{30} =4/5 $
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