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(4x+2)(3x-7)=103
We move all terms to the left:
(4x+2)(3x-7)-(103)=0
We multiply parentheses ..
(+12x^2-28x+6x-14)-103=0
We get rid of parentheses
12x^2-28x+6x-14-103=0
We add all the numbers together, and all the variables
12x^2-22x-117=0
a = 12; b = -22; c = -117;
Δ = b2-4ac
Δ = -222-4·12·(-117)
Δ = 6100
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{6100}=\sqrt{100*61}=\sqrt{100}*\sqrt{61}=10\sqrt{61}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-10\sqrt{61}}{2*12}=\frac{22-10\sqrt{61}}{24} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+10\sqrt{61}}{2*12}=\frac{22+10\sqrt{61}}{24} $
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