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(4x-5)(3x+5)=40
We move all terms to the left:
(4x-5)(3x+5)-(40)=0
We multiply parentheses ..
(+12x^2+20x-15x-25)-40=0
We get rid of parentheses
12x^2+20x-15x-25-40=0
We add all the numbers together, and all the variables
12x^2+5x-65=0
a = 12; b = 5; c = -65;
Δ = b2-4ac
Δ = 52-4·12·(-65)
Δ = 3145
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-\sqrt{3145}}{2*12}=\frac{-5-\sqrt{3145}}{24} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{3145}}{2*12}=\frac{-5+\sqrt{3145}}{24} $
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