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(6x-15)x=96
We move all terms to the left:
(6x-15)x-(96)=0
We multiply parentheses
6x^2-15x-96=0
a = 6; b = -15; c = -96;
Δ = b2-4ac
Δ = -152-4·6·(-96)
Δ = 2529
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2529}=\sqrt{9*281}=\sqrt{9}*\sqrt{281}=3\sqrt{281}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-3\sqrt{281}}{2*6}=\frac{15-3\sqrt{281}}{12} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+3\sqrt{281}}{2*6}=\frac{15+3\sqrt{281}}{12} $
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