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(G)=3G^2-14G-5
We move all terms to the left:
(G)-(3G^2-14G-5)=0
We get rid of parentheses
-3G^2+G+14G+5=0
We add all the numbers together, and all the variables
-3G^2+15G+5=0
a = -3; b = 15; c = +5;
Δ = b2-4ac
Δ = 152-4·(-3)·5
Δ = 285
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$G_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$G_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$G_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-\sqrt{285}}{2*-3}=\frac{-15-\sqrt{285}}{-6} $$G_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+\sqrt{285}}{2*-3}=\frac{-15+\sqrt{285}}{-6} $
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