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24x^2+33x=516
We move all terms to the left:
24x^2+33x-(516)=0
a = 24; b = 33; c = -516;
Δ = b2-4ac
Δ = 332-4·24·(-516)
Δ = 50625
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{50625}=225$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(33)-225}{2*24}=\frac{-258}{48} =-5+3/8 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(33)+225}{2*24}=\frac{192}{48} =4 $
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