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2r^2+3=-7r
We move all terms to the left:
2r^2+3-(-7r)=0
We get rid of parentheses
2r^2+7r+3=0
a = 2; b = 7; c = +3;
Δ = b2-4ac
Δ = 72-4·2·3
Δ = 25
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}$$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{25}=5$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-5}{2*2}=\frac{-12}{4} =-3 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+5}{2*2}=\frac{-2}{4} =-1/2 $
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