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(7x+3)(4x-5)=0
We multiply parentheses ..
(+28x^2-35x+12x-15)=0
We get rid of parentheses
28x^2-35x+12x-15=0
We add all the numbers together, and all the variables
28x^2-23x-15=0
a = 28; b = -23; c = -15;
Δ = b2-4ac
Δ = -232-4·28·(-15)
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-23)-47}{2*28}=\frac{-24}{56} =-3/7 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-23)+47}{2*28}=\frac{70}{56} =1+1/4 $
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