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(3x+16)(x-6)=0
We multiply parentheses ..
(+3x^2-18x+16x-96)=0
We get rid of parentheses
3x^2-18x+16x-96=0
We add all the numbers together, and all the variables
3x^2-2x-96=0
a = 3; b = -2; c = -96;
Δ = b2-4ac
Δ = -22-4·3·(-96)
Δ = 1156
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{1156}=34$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-34}{2*3}=\frac{-32}{6} =-5+1/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+34}{2*3}=\frac{36}{6} =6 $
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