What are the solutions to the equation t^2=-63t+650 ?

The solutions to the equation t^2=-63t+650 are t=(-63+sqrt(6569))/2 ,t=(-63-sqrt(6569))/2

Find the zeros of t^2=-63t+650

The zeros of t^2=-63t+650 are t=(-63+sqrt(6569))/2 ,t=(-63-sqrt(6569))/2

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t^{2} = - 63 t + 650

t = \frac{- 63 + 6569}{2}, t = \frac{- 63 - 6569}{2}

Decimal Notation

t = 9.02468 \dots, t = - 72.02468 \dots

Solution steps

t^{2} = - 63 t + 650

Move

650

to the left side

t^{2} - 650 = - 63 t

Move

63 t

to the left side

t^{2} - 650 + 63 t = 0

Write in the standard form

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

t^{2} + 63 t - 650 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 63 \pm 6 3 ^{2} - 4 \cdot 1 \cdot ( - 650 )}{2 \cdot 1}

6 3^{2} - 4 \cdot 1 \cdot (- 650) = 6569

t_{1, 2} = \frac{- 63 \pm 6569}{2 \cdot 1}

Separate the solutions

t_{1} = \frac{- 63 + 6569}{2 \cdot 1}, t_{2} = \frac{- 63 - 6569}{2 \cdot 1}

t = \frac{- 63 + 6569}{2 \cdot 1} : \frac{- 63 + 6569}{2}

t = \frac{- 63 - 6569}{2 \cdot 1} : \frac{- 63 - 6569}{2}

The solutions to the quadratic equation are:

t = \frac{- 63 + 6569}{2}, t = \frac{- 63 - 6569}{2}

3 r^{2} + 8 r + 6 = 0

x, y^{1} + 3 x^{2} y = x^{2}

3 (g - 4)^{2} = 27

70 x^{2} + 120 x = 440

2 x^{2} - 14 x + 10 = 26

What are the solutions to the equation t^2=-63t+650 ?
The solutions to the equation t^2=-63t+650 are t=(-63+sqrt(6569))/2 ,t=(-63-sqrt(6569))/2
Find the zeros of t^2=-63t+650
The zeros of t^2=-63t+650 are t=(-63+sqrt(6569))/2 ,t=(-63-sqrt(6569))/2