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x(3x+4)=22
We move all terms to the left:
x(3x+4)-(22)=0
We multiply parentheses
3x^2+4x-22=0
a = 3; b = 4; c = -22;
Δ = b2-4ac
Δ = 42-4·3·(-22)
Δ = 280
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{280}=\sqrt{4*70}=\sqrt{4}*\sqrt{70}=2\sqrt{70}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{70}}{2*3}=\frac{-4-2\sqrt{70}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{70}}{2*3}=\frac{-4+2\sqrt{70}}{6} $
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