What are the solutions to the equation v^2+11v-26=0 ?

The solutions to the equation v^2+11v-26=0 are v=2,v=-13

Find the zeros of v^2+11v-26=0

The zeros of v^2+11v-26=0 are v=2,v=-13

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v^{2} + 11 v - 26 = 0

v = 2, v = - 13

Solution steps

v^{2} + 11 v - 26 = 0

Solve with the quadratic formula

v_{1, 2} = \frac{- 11 \pm 1 1 ^{2} - 4 \cdot 1 \cdot ( - 26 )}{2 \cdot 1}

1 1^{2} - 4 \cdot 1 \cdot (- 26) = 15

v_{1, 2} = \frac{- 11 \pm 15}{2 \cdot 1}

Separate the solutions

v_{1} = \frac{- 11 + 15}{2 \cdot 1}, v_{2} = \frac{- 11 - 15}{2 \cdot 1}

v = \frac{- 11 + 15}{2 \cdot 1} : 2

v = \frac{- 11 - 15}{2 \cdot 1} : - 13

The solutions to the quadratic equation are:

v = 2, v = - 13

159

\frac{1}{2} x + 8 \leq 10

∣ 2 x + 1 ∣ + 1 \leq 6

\frac{3}{2} y - 1 = 11

\frac{( 8 + 2 i ) - ( 1 - i )}{( 2 + i ) ^{2}}

What are the solutions to the equation v^2+11v-26=0 ?
The solutions to the equation v^2+11v-26=0 are v=2,v=-13
Find the zeros of v^2+11v-26=0
The zeros of v^2+11v-26=0 are v=2,v=-13