What are the solutions to the equation t^2+2t-19=5 ?

The solutions to the equation t^2+2t-19=5 are t=4,t=-6

Find the zeros of t^2+2t-19=5

The zeros of t^2+2t-19=5 are t=4,t=-6

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t^{2} + 2 t - 19 = 5

t = 4, t = - 6

Solution steps

t^{2} + 2 t - 19 = 5

Move

5

to the left side

t^{2} + 2 t - 24 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 1 \cdot ( - 24 )}{2 \cdot 1}

2^{2} - 4 \cdot 1 \cdot (- 24) = 10

t_{1, 2} = \frac{- 2 \pm 10}{2 \cdot 1}

Separate the solutions

t_{1} = \frac{- 2 + 10}{2 \cdot 1}, t_{2} = \frac{- 2 - 10}{2 \cdot 1}

t = \frac{- 2 + 10}{2 \cdot 1} : 4

t = \frac{- 2 - 10}{2 \cdot 1} : - 6

The solutions to the quadratic equation are:

t = 4, t = - 6

2 x^{2} - 4 x + 15

u^{2} - 3 u - 10 = 0

6 y + 3 = - 4 y^{2}

(2 x + 7)^{2} = 16 x^{2}

x^{2} + x - 1 = 11

What are the solutions to the equation t^2+2t-19=5 ?
The solutions to the equation t^2+2t-19=5 are t=4,t=-6
Find the zeros of t^2+2t-19=5
The zeros of t^2+2t-19=5 are t=4,t=-6