3b^2=12b+63

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Solution for 3b^2=12b+63 equation:



3b^2=12b+63
We move all terms to the left:
3b^2-(12b+63)=0
We get rid of parentheses
3b^2-12b-63=0
a = 3; b = -12; c = -63;
Δ = b2-4ac
Δ = -122-4·3·(-63)
Δ = 900
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{900}=30$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-30}{2*3}=\frac{-18}{6} =-3 $
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+30}{2*3}=\frac{42}{6} =7 $

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