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6x^2+36x=15x
We move all terms to the left:
6x^2+36x-(15x)=0
We add all the numbers together, and all the variables
6x^2+21x=0
a = 6; b = 21; c = 0;
Δ = b2-4ac
Δ = 212-4·6·0
Δ = 441
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{441}=21$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(21)-21}{2*6}=\frac{-42}{12} =-3+1/2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+21}{2*6}=\frac{0}{12} =0 $
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