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5b^2+4=-12b
We move all terms to the left:
5b^2+4-(-12b)=0
We get rid of parentheses
5b^2+12b+4=0
a = 5; b = 12; c = +4;
Δ = b2-4ac
Δ = 122-4·5·4
Δ = 64
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{64}=8$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-8}{2*5}=\frac{-20}{10} =-2 $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+8}{2*5}=\frac{-4}{10} =-2/5 $
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