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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^2-47x+28=0
a = 15; b = -47; c = +28;
Δ = b2-4ac
Δ = -472-4·15·28
Δ = 529
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{529}=23$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-47)-23}{2*15}=\frac{24}{30} =4/5 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-47)+23}{2*15}=\frac{70}{30} =2+1/3 $
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