What are the solutions to the equation t^2-5t=-4 ?

The solutions to the equation t^2-5t=-4 are t=4,t=1

Find the zeros of t^2-5t=-4

The zeros of t^2-5t=-4 are t=4,t=1

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t^{2} - 5 t = - 4

t = 4, t = 1

Solution steps

t^{2} - 5 t = - 4

Move

4

to the left side

t^{2} - 5 t + 4 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 1 \cdot 4}{2 \cdot 1}

(- 5)^{2} - 4 \cdot 1 \cdot 4 = 3

t_{1, 2} = \frac{- ( - 5 ) \pm 3}{2 \cdot 1}

Separate the solutions

t_{1} = \frac{- ( - 5 ) + 3}{2 \cdot 1}, t_{2} = \frac{- ( - 5 ) - 3}{2 \cdot 1}

t = \frac{- ( - 5 ) + 3}{2 \cdot 1} : 4

t = \frac{- ( - 5 ) - 3}{2 \cdot 1} : 1

The solutions to the quadratic equation are:

t = 4, t = 1

9 (\frac{10}{13} - \frac{1}{3})

- 2 ∣ x - 3 ∣ \geq - 4

\frac{3 x - 1}{4} = 5

- 12 x \leq - 94 + 85 x

3 z^{2} - 2 z - 5

What are the solutions to the equation t^2-5t=-4 ?
The solutions to the equation t^2-5t=-4 are t=4,t=1
Find the zeros of t^2-5t=-4
The zeros of t^2-5t=-4 are t=4,t=1