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9x^2+48x+64=
We move all terms to the left:
9x^2+48x+64-()=0
We add all the numbers together, and all the variables
9x^2+48x=0
a = 9; b = 48; c = 0;
Δ = b2-4ac
Δ = 482-4·9·0
Δ = 2304
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{2304}=48$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(48)-48}{2*9}=\frac{-96}{18} =-5+1/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(48)+48}{2*9}=\frac{0}{18} =0 $
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