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2(3x^2+10x-3)=0
We multiply parentheses
6x^2+20x-6=0
a = 6; b = 20; c = -6;
Δ = b2-4ac
Δ = 202-4·6·(-6)
Δ = 544
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{544}=\sqrt{16*34}=\sqrt{16}*\sqrt{34}=4\sqrt{34}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4\sqrt{34}}{2*6}=\frac{-20-4\sqrt{34}}{12} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4\sqrt{34}}{2*6}=\frac{-20+4\sqrt{34}}{12} $
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