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12a^2+10a=14
We move all terms to the left:
12a^2+10a-(14)=0
a = 12; b = 10; c = -14;
Δ = b2-4ac
Δ = 102-4·12·(-14)
Δ = 772
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{772}=\sqrt{4*193}=\sqrt{4}*\sqrt{193}=2\sqrt{193}$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{193}}{2*12}=\frac{-10-2\sqrt{193}}{24} $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{193}}{2*12}=\frac{-10+2\sqrt{193}}{24} $
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