What are the solutions to the equation 18t+2t^2-180=0 ?

The solutions to the equation 18t+2t^2-180=0 are t=6,t=-15

Find the zeros of 18t+2t^2-180=0

The zeros of 18t+2t^2-180=0 are t=6,t=-15

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18 t + 2 t^{2} - 180 = 0

t = 6, t = - 15

Solution steps

18 t + 2 t^{2} - 180 = 0

Write in the standard form

a x^{2} + b x + c = 0

2 t^{2} + 18 t - 180 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 18 \pm 1 8 ^{2} - 4 \cdot 2 ( - 180 )}{2 \cdot 2}

1 8^{2} - 4 \cdot 2 (- 180) = 42

t_{1, 2} = \frac{- 18 \pm 42}{2 \cdot 2}

Separate the solutions

t_{1} = \frac{- 18 + 42}{2 \cdot 2}, t_{2} = \frac{- 18 - 42}{2 \cdot 2}

t = \frac{- 18 + 42}{2 \cdot 2} : 6

t = \frac{- 18 - 42}{2 \cdot 2} : - 15

The solutions to the quadratic equation are:

t = 6, t = - 15

n^{2} = 6 + n

3 (x + 4)^{2} = 25

x^{2} + (2 x + 1)^{2} = 10

5 n^{2} = 6 n - 1

- 3 = g^{2} - 4 g

What are the solutions to the equation 18t+2t^2-180=0 ?
The solutions to the equation 18t+2t^2-180=0 are t=6,t=-15
Find the zeros of 18t+2t^2-180=0
The zeros of 18t+2t^2-180=0 are t=6,t=-15