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2x(3x+19)=114
We move all terms to the left:
2x(3x+19)-(114)=0
We multiply parentheses
6x^2+38x-114=0
a = 6; b = 38; c = -114;
Δ = b2-4ac
Δ = 382-4·6·(-114)
Δ = 4180
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{4180}=\sqrt{4*1045}=\sqrt{4}*\sqrt{1045}=2\sqrt{1045}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(38)-2\sqrt{1045}}{2*6}=\frac{-38-2\sqrt{1045}}{12} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(38)+2\sqrt{1045}}{2*6}=\frac{-38+2\sqrt{1045}}{12} $
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