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(7x-4)(x-3)=3x+12
We move all terms to the left:
(7x-4)(x-3)-(3x+12)=0
We get rid of parentheses
(7x-4)(x-3)-3x-12=0
We multiply parentheses ..
(+7x^2-21x-4x+12)-3x-12=0
We get rid of parentheses
7x^2-21x-4x-3x+12-12=0
We add all the numbers together, and all the variables
7x^2-28x=0
a = 7; b = -28; c = 0;
Δ = b2-4ac
Δ = -282-4·7·0
Δ = 784
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{784}=28$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-28}{2*7}=\frac{0}{14} =0 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+28}{2*7}=\frac{56}{14} =4 $
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