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

The solutions to the equation -r^2-r+12=0 are r=-4,r=3

Find the zeros of -r^2-r+12=0

The zeros of -r^2-r+12=0 are r=-4,r=3

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- r^{2} - r + 12 = 0

r = - 4, r = 3

Solution steps

- r^{2} - r + 12 = 0

Solve with the quadratic formula

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

(- 1)^{2} - 4 (- 1) \cdot 12 = 7

r_{1, 2} = \frac{- ( - 1 ) \pm 7}{2 ( - 1 )}

Separate the solutions

r_{1} = \frac{- ( - 1 ) + 7}{2 ( - 1 )}, r_{2} = \frac{- ( - 1 ) - 7}{2 ( - 1 )}

r = \frac{- ( - 1 ) + 7}{2 ( - 1 )} : - 4

r = \frac{- ( - 1 ) - 7}{2 ( - 1 )} : 3

The solutions to the quadratic equation are:

r = - 4, r = 3

0 = 3.7 (x - 8)^{2} - 5

x^{2} + \frac{2}{3} x = 0

- 3 t (t + 2) = - 7

2 x^{2} + 7 x + 7

x^{2} - 10 x + 52 = 0

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