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x(3x+48)=1260
We move all terms to the left:
x(3x+48)-(1260)=0
We multiply parentheses
3x^2+48x-1260=0
a = 3; b = 48; c = -1260;
Δ = b2-4ac
Δ = 482-4·3·(-1260)
Δ = 17424
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{17424}=132$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(48)-132}{2*3}=\frac{-180}{6} =-30 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(48)+132}{2*3}=\frac{84}{6} =14 $
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