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3(6x-4)+5x(3x+2)=10
We move all terms to the left:
3(6x-4)+5x(3x+2)-(10)=0
We multiply parentheses
15x^2+18x+10x-12-10=0
We add all the numbers together, and all the variables
15x^2+28x-22=0
a = 15; b = 28; c = -22;
Δ = b2-4ac
Δ = 282-4·15·(-22)
Δ = 2104
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2104}=\sqrt{4*526}=\sqrt{4}*\sqrt{526}=2\sqrt{526}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-2\sqrt{526}}{2*15}=\frac{-28-2\sqrt{526}}{30} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+2\sqrt{526}}{2*15}=\frac{-28+2\sqrt{526}}{30} $
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