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(4x+2)(3x-7)=100
We move all terms to the left:
(4x+2)(3x-7)-(100)=0
We multiply parentheses ..
(+12x^2-28x+6x-14)-100=0
We get rid of parentheses
12x^2-28x+6x-14-100=0
We add all the numbers together, and all the variables
12x^2-22x-114=0
a = 12; b = -22; c = -114;
Δ = b2-4ac
Δ = -222-4·12·(-114)
Δ = 5956
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{5956}=\sqrt{4*1489}=\sqrt{4}*\sqrt{1489}=2\sqrt{1489}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-2\sqrt{1489}}{2*12}=\frac{22-2\sqrt{1489}}{24} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+2\sqrt{1489}}{2*12}=\frac{22+2\sqrt{1489}}{24} $
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