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x=3x^2-15x
We move all terms to the left:
x-(3x^2-15x)=0
We get rid of parentheses
-3x^2+x+15x=0
We add all the numbers together, and all the variables
-3x^2+16x=0
a = -3; b = 16; c = 0;
Δ = b2-4ac
Δ = 162-4·(-3)·0
Δ = 256
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{256}=16$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-16}{2*-3}=\frac{-32}{-6} =5+1/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+16}{2*-3}=\frac{0}{-6} =0 $
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