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3x^2+18=15x
We move all terms to the left:
3x^2+18-(15x)=0
a = 3; b = -15; c = +18;
Δ = b2-4ac
Δ = -152-4·3·18
Δ = 9
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{9}=3$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-3}{2*3}=\frac{12}{6} =2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+3}{2*3}=\frac{18}{6} =3 $
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