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(6x-3)(4x-3)=8x+14
We move all terms to the left:
(6x-3)(4x-3)-(8x+14)=0
We get rid of parentheses
(6x-3)(4x-3)-8x-14=0
We multiply parentheses ..
(+24x^2-18x-12x+9)-8x-14=0
We get rid of parentheses
24x^2-18x-12x-8x+9-14=0
We add all the numbers together, and all the variables
24x^2-38x-5=0
a = 24; b = -38; c = -5;
Δ = b2-4ac
Δ = -382-4·24·(-5)
Δ = 1924
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1924}=\sqrt{4*481}=\sqrt{4}*\sqrt{481}=2\sqrt{481}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-38)-2\sqrt{481}}{2*24}=\frac{38-2\sqrt{481}}{48} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-38)+2\sqrt{481}}{2*24}=\frac{38+2\sqrt{481}}{48} $
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