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4r^2+-15r+9=0
We add all the numbers together, and all the variables
4r^2-15r=0
a = 4; b = -15; c = 0;
Δ = b2-4ac
Δ = -152-4·4·0
Δ = 225
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{225}=15$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-15}{2*4}=\frac{0}{8} =0 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+15}{2*4}=\frac{30}{8} =3+3/4 $
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