What are the solutions to the equation t^2+t=20 ?

The solutions to the equation t^2+t=20 are t=4,t=-5

Find the zeros of t^2+t=20

The zeros of t^2+t=20 are t=4,t=-5

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t^{2} + t = 20

t = 4, t = - 5

Solution steps

t^{2} + t = 20

Move

20

to the left side

t^{2} + t - 20 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 1 \cdot ( - 20 )}{2 \cdot 1}

1^{2} - 4 \cdot 1 \cdot (- 20) = 9

t_{1, 2} = \frac{- 1 \pm 9}{2 \cdot 1}

Separate the solutions

t_{1} = \frac{- 1 + 9}{2 \cdot 1}, t_{2} = \frac{- 1 - 9}{2 \cdot 1}

t = \frac{- 1 + 9}{2 \cdot 1} : 4

t = \frac{- 1 - 9}{2 \cdot 1} : - 5

The solutions to the quadratic equation are:

t = 4, t = - 5

(3 u)^{2}

v = 32 (2.9)

49 a^{2} - b^{2}

ln (0.25)

3 x - (3 + 2 x^{3}) + 5 x = 2 (3 x - 13)

What are the solutions to the equation t^2+t=20 ?
The solutions to the equation t^2+t=20 are t=4,t=-5
Find the zeros of t^2+t=20
The zeros of t^2+t=20 are t=4,t=-5