What are the solutions to the equation r^2+4r-12=0 ?

The solutions to the equation r^2+4r-12=0 are r=2,r=-6

Find the zeros of r^2+4r-12=0

The zeros of r^2+4r-12=0 are r=2,r=-6

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r^{2} + 4 r - 12 = 0

r = 2, r = - 6

Solution steps

r^{2} + 4 r - 12 = 0

Solve with the quadratic formula

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

4^{2} - 4 \cdot 1 \cdot (- 12) = 8

r_{1, 2} = \frac{- 4 \pm 8}{2 \cdot 1}

Separate the solutions

r_{1} = \frac{- 4 + 8}{2 \cdot 1}, r_{2} = \frac{- 4 - 8}{2 \cdot 1}

r = \frac{- 4 + 8}{2 \cdot 1} : 2

r = \frac{- 4 - 8}{2 \cdot 1} : - 6

The solutions to the quadratic equation are:

r = 2, r = - 6

2 x^{2} + 15 x - 8 = 0

(4 a + 3 b)^{2} - (3 a - 2 b)^{2}

\frac{x + 5}{5} = \frac{9}{8}

\frac{8 a ^{3} - 10 a}{20 a ^{3} - 25 a}

3 x - 6 = - 9 x + 4

What are the solutions to the equation r^2+4r-12=0 ?
The solutions to the equation r^2+4r-12=0 are r=2,r=-6
Find the zeros of r^2+4r-12=0
The zeros of r^2+4r-12=0 are r=2,r=-6