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(5x+4)(3x-7)=0
We multiply parentheses ..
(+15x^2-35x+12x-28)=0
We get rid of parentheses
15x^2-35x+12x-28=0
We add all the numbers together, and all the variables
15x^2-23x-28=0
a = 15; b = -23; c = -28;
Δ = b2-4ac
Δ = -232-4·15·(-28)
Δ = 2209
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{2209}=47$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-23)-47}{2*15}=\frac{-24}{30} =-4/5 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-23)+47}{2*15}=\frac{70}{30} =2+1/3 $
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