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16x-3x^2=5
We move all terms to the left:
16x-3x^2-(5)=0
a = -3; b = 16; c = -5;
Δ = b2-4ac
Δ = 162-4·(-3)·(-5)
Δ = 196
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{196}=14$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-14}{2*-3}=\frac{-30}{-6} =+5 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+14}{2*-3}=\frac{-2}{-6} =1/3 $
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