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

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

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

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

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

t = 22, t = 48

Solution steps

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

Switch sides

- 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

a^{2} - a = 6

4 x^{2} - 6 x - 9 = 0

x^{2} + \frac{1}{2} x + \frac{1}{2} = 0

4 x^{2} - 40 x + 96 = 0

x^{2} + (- 2 x - 38)^{2} + 24 x + 8 (- 2 x - 38) + 120 = 0

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