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3r^2-6r-7=5
We move all terms to the left:
3r^2-6r-7-(5)=0
We add all the numbers together, and all the variables
3r^2-6r-12=0
a = 3; b = -6; c = -12;
Δ = b2-4ac
Δ = -62-4·3·(-12)
Δ = 180
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{180}=\sqrt{36*5}=\sqrt{36}*\sqrt{5}=6\sqrt{5}$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-6\sqrt{5}}{2*3}=\frac{6-6\sqrt{5}}{6} $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+6\sqrt{5}}{2*3}=\frac{6+6\sqrt{5}}{6} $
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