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2r^2+5r+3=9
We move all terms to the left:
2r^2+5r+3-(9)=0
We add all the numbers together, and all the variables
2r^2+5r-6=0
a = 2; b = 5; c = -6;
Δ = b2-4ac
Δ = 52-4·2·(-6)
Δ = 73
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}$$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-\sqrt{73}}{2*2}=\frac{-5-\sqrt{73}}{4} $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{73}}{2*2}=\frac{-5+\sqrt{73}}{4} $
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