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(2x+3)(3x-7)=40
We move all terms to the left:
(2x+3)(3x-7)-(40)=0
We multiply parentheses ..
(+6x^2-14x+9x-21)-40=0
We get rid of parentheses
6x^2-14x+9x-21-40=0
We add all the numbers together, and all the variables
6x^2-5x-61=0
a = 6; b = -5; c = -61;
Δ = b2-4ac
Δ = -52-4·6·(-61)
Δ = 1489
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{1489}}{2*6}=\frac{5-\sqrt{1489}}{12} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+\sqrt{1489}}{2*6}=\frac{5+\sqrt{1489}}{12} $
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