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(2x+4)(3x-13)=62
We move all terms to the left:
(2x+4)(3x-13)-(62)=0
We multiply parentheses ..
(+6x^2-26x+12x-52)-62=0
We get rid of parentheses
6x^2-26x+12x-52-62=0
We add all the numbers together, and all the variables
6x^2-14x-114=0
a = 6; b = -14; c = -114;
Δ = b2-4ac
Δ = -142-4·6·(-114)
Δ = 2932
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2932}=\sqrt{4*733}=\sqrt{4}*\sqrt{733}=2\sqrt{733}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-2\sqrt{733}}{2*6}=\frac{14-2\sqrt{733}}{12} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+2\sqrt{733}}{2*6}=\frac{14+2\sqrt{733}}{12} $
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