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9x^2=21x
We move all terms to the left:
9x^2-(21x)=0
a = 9; b = -21; c = 0;
Δ = b2-4ac
Δ = -212-4·9·0
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-21)-21}{2*9}=\frac{0}{18} =0 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-21)+21}{2*9}=\frac{42}{18} =2+1/3 $
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