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3x^2-16x-3=-9x+3
We move all terms to the left:
3x^2-16x-3-(-9x+3)=0
We get rid of parentheses
3x^2-16x+9x-3-3=0
We add all the numbers together, and all the variables
3x^2-7x-6=0
a = 3; b = -7; c = -6;
Δ = b2-4ac
Δ = -72-4·3·(-6)
Δ = 121
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{121}=11$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-11}{2*3}=\frac{-4}{6} =-2/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+11}{2*3}=\frac{18}{6} =3 $
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