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3b^2-20=7b
We move all terms to the left:
3b^2-20-(7b)=0
a = 3; b = -7; c = -20;
Δ = b2-4ac
Δ = -72-4·3·(-20)
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{289}=17$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-17}{2*3}=\frac{-10}{6} =-1+2/3 $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+17}{2*3}=\frac{24}{6} =4 $
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