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3x^2+17x-28=132
We move all terms to the left:
3x^2+17x-28-(132)=0
We add all the numbers together, and all the variables
3x^2+17x-160=0
a = 3; b = 17; c = -160;
Δ = b2-4ac
Δ = 172-4·3·(-160)
Δ = 2209
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{2209}=47$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(17)-47}{2*3}=\frac{-64}{6} =-10+2/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+47}{2*3}=\frac{30}{6} =5 $
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