What are the solutions to the equation -t^2+70t+460=1516 ?

The solutions to the equation -t^2+70t+460=1516 are t=22,t=48

Find the zeros of -t^2+70t+460=1516

The zeros of -t^2+70t+460=1516 are t=22,t=48

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- t^{2} + 70 t + 460 = 1516

t = 22, t = 48

Solution steps

- t^{2} + 70 t + 460 = 1516

Move

1516

to the left side

- t^{2} + 70 t - 1056 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 70 \pm 7 0 ^{2} - 4 ( - 1 ) ( - 1056 )}{2 ( - 1 )}

7 0^{2} - 4 (- 1) (- 1056) = 26

t_{1, 2} = \frac{- 70 \pm 26}{2 ( - 1 )}

Separate the solutions

t_{1} = \frac{- 70 + 26}{2 ( - 1 )}, t_{2} = \frac{- 70 - 26}{2 ( - 1 )}

t = \frac{- 70 + 26}{2 ( - 1 )} : 22

t = \frac{- 70 - 26}{2 ( - 1 )} : 48

The solutions to the quadratic equation are:

t = 22, t = 48

10 x - x^{2} = 0

3 (x + 2)^{2} - 11 = 25

10 x^{2} - 8 x - 2 = 0

10 x^{2} - 9 x = - 5 x^{2} + 11 x

j (x) = x^{2}

What are the solutions to the equation -t^2+70t+460=1516 ?
The solutions to the equation -t^2+70t+460=1516 are t=22,t=48
Find the zeros of -t^2+70t+460=1516
The zeros of -t^2+70t+460=1516 are t=22,t=48