What are the solutions to the equation 12r^2-11r-15=0 ?

The solutions to the equation 12r^2-11r-15=0 are r= 5/3 ,r=-3/4

Find the zeros of 12r^2-11r-15=0

The zeros of 12r^2-11r-15=0 are r= 5/3 ,r=-3/4

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12 r^{2} - 11 r - 15 = 0

r = \frac{5}{3}, r = - \frac{3}{4}

Solution steps

12 r^{2} - 11 r - 15 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- ( - 11 ) \pm ( - 11 ) ^{2} - 4 \cdot 12 ( - 15 )}{2 \cdot 12}

(- 11)^{2} - 4 \cdot 12 (- 15) = 29

r_{1, 2} = \frac{- ( - 11 ) \pm 29}{2 \cdot 12}

Separate the solutions

r_{1} = \frac{- ( - 11 ) + 29}{2 \cdot 12}, r_{2} = \frac{- ( - 11 ) - 29}{2 \cdot 12}

r = \frac{- ( - 11 ) + 29}{2 \cdot 12} : \frac{5}{3}

r = \frac{- ( - 11 ) - 29}{2 \cdot 12} : - \frac{3}{4}

The solutions to the quadratic equation are:

r = \frac{5}{3}, r = - \frac{3}{4}

s^{2} + 3 s + 5 = 0

6 \cdot 0.5

\frac{3}{5} x - 1 = 2

g (f (2))

13 y^{2} + 10 y - 3 = 5 y^{2}

What are the solutions to the equation 12r^2-11r-15=0 ?
The solutions to the equation 12r^2-11r-15=0 are r= 5/3 ,r=-3/4
Find the zeros of 12r^2-11r-15=0
The zeros of 12r^2-11r-15=0 are r= 5/3 ,r=-3/4