What are the solutions to the equation t^2+3t-28=0 ?

The solutions to the equation t^2+3t-28=0 are t=4,t=-7

Find the zeros of t^2+3t-28=0

The zeros of t^2+3t-28=0 are t=4,t=-7

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t^{2} + 3 t - 28 = 0

t = 4, t = - 7

Solution steps

t^{2} + 3 t - 28 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 1 \cdot ( - 28 )}{2 \cdot 1}

3^{2} - 4 \cdot 1 \cdot (- 28) = 11

t_{1, 2} = \frac{- 3 \pm 11}{2 \cdot 1}

Separate the solutions

t_{1} = \frac{- 3 + 11}{2 \cdot 1}, t_{2} = \frac{- 3 - 11}{2 \cdot 1}

t = \frac{- 3 + 11}{2 \cdot 1} : 4

t = \frac{- 3 - 11}{2 \cdot 1} : - 7

The solutions to the quadratic equation are:

t = 4, t = - 7

40 x \geq 25 x + 600

∣ - 14 ∣

- x^{4} + x^{3} + 4 x^{2} - 4 x

0 = x^{2} - 6 x - 7

∣ 7 x - 1 ∣ + 4 \geq 9

What are the solutions to the equation t^2+3t-28=0 ?
The solutions to the equation t^2+3t-28=0 are t=4,t=-7
Find the zeros of t^2+3t-28=0
The zeros of t^2+3t-28=0 are t=4,t=-7