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(9x-4)(3x+7)=0
We multiply parentheses ..
(+27x^2+63x-12x-28)=0
We get rid of parentheses
27x^2+63x-12x-28=0
We add all the numbers together, and all the variables
27x^2+51x-28=0
a = 27; b = 51; c = -28;
Δ = b2-4ac
Δ = 512-4·27·(-28)
Δ = 5625
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{5625}=75$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(51)-75}{2*27}=\frac{-126}{54} =-2+1/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(51)+75}{2*27}=\frac{24}{54} =4/9 $
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