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(4x-5)(3x+5)+40=180
We move all terms to the left:
(4x-5)(3x+5)+40-(180)=0
We add all the numbers together, and all the variables
(4x-5)(3x+5)-140=0
We multiply parentheses ..
(+12x^2+20x-15x-25)-140=0
We get rid of parentheses
12x^2+20x-15x-25-140=0
We add all the numbers together, and all the variables
12x^2+5x-165=0
a = 12; b = 5; c = -165;
Δ = b2-4ac
Δ = 52-4·12·(-165)
Δ = 7945
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{7945}}{2*12}=\frac{-5-\sqrt{7945}}{24} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{7945}}{2*12}=\frac{-5+\sqrt{7945}}{24} $
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