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x(3x+10)=30
We move all terms to the left:
x(3x+10)-(30)=0
We multiply parentheses
3x^2+10x-30=0
a = 3; b = 10; c = -30;
Δ = b2-4ac
Δ = 102-4·3·(-30)
Δ = 460
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{460}=\sqrt{4*115}=\sqrt{4}*\sqrt{115}=2\sqrt{115}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{115}}{2*3}=\frac{-10-2\sqrt{115}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{115}}{2*3}=\frac{-10+2\sqrt{115}}{6} $
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