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(5x-4)(3x+8)=0
We multiply parentheses ..
(+15x^2+40x-12x-32)=0
We get rid of parentheses
15x^2+40x-12x-32=0
We add all the numbers together, and all the variables
15x^2+28x-32=0
a = 15; b = 28; c = -32;
Δ = b2-4ac
Δ = 282-4·15·(-32)
Δ = 2704
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{2704}=52$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-52}{2*15}=\frac{-80}{30} =-2+2/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+52}{2*15}=\frac{24}{30} =4/5 $
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