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4r^2-7=-3r
We move all terms to the left:
4r^2-7-(-3r)=0
We get rid of parentheses
4r^2+3r-7=0
a = 4; b = 3; c = -7;
Δ = b2-4ac
Δ = 32-4·4·(-7)
Δ = 121
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{121}=11$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-11}{2*4}=\frac{-14}{8} =-1+3/4 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+11}{2*4}=\frac{8}{8} =1 $
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