What are the solutions to the equation t^2-7t+16=t ?

The solutions to the equation t^2-7t+16=t are t=4

Find the zeros of t^2-7t+16=t

The zeros of t^2-7t+16=t are t=4

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t^{2} - 7 t + 16 = t

t = 4

Solution steps

t^{2} - 7 t + 16 = t

Move

t

to the left side

t^{2} - 8 t + 16 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 \cdot 1 \cdot 16}{2 \cdot 1}

(- 8)^{2} - 4 \cdot 1 \cdot 16 = 0

t_{1, 2} = \frac{- ( - 8 ) \pm 0}{2 \cdot 1}

t = \frac{- ( - 8 )}{2 \cdot 1}

\frac{- ( - 8 )}{2 \cdot 1} = 4

t = 4

The solution to the quadratic equation is:

t = 4

\frac{1}{8} \cdot 7

- 2 ∣ x + 8 ∣ + 7 = - 21

x^{2} - 11 x + 28 > 0

\frac{( d ^{2} y )}{d x ^{2}}

(- 36)^{\frac{1}{2}}

What are the solutions to the equation t^2-7t+16=t ?
The solutions to the equation t^2-7t+16=t are t=4
Find the zeros of t^2-7t+16=t
The zeros of t^2-7t+16=t are t=4