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16x-48-x^2=0
We add all the numbers together, and all the variables
-1x^2+16x-48=0
a = -1; b = 16; c = -48;
Δ = b2-4ac
Δ = 162-4·(-1)·(-48)
Δ = 64
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{64}=8$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-8}{2*-1}=\frac{-24}{-2} =+12 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+8}{2*-1}=\frac{-8}{-2} =+4 $
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