What are the solutions to the equation t^2+2t+6=7t+2 ?

The solutions to the equation t^2+2t+6=7t+2 are t=4,t=1

Find the zeros of t^2+2t+6=7t+2

The zeros of t^2+2t+6=7t+2 are t=4,t=1

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t^{2} + 2 t + 6 = 7 t + 2

t = 4, t = 1

Solution steps

t^{2} + 2 t + 6 = 7 t + 2

Move

2

to the left side

t^{2} + 2 t + 4 = 7 t

Move

7 t

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

x^{2} - 16 x - 36 = 0

k^{2} - 2 k = - 2

0.75 \frac{π d ^{2}}{4} = 6573.35

\frac{1}{2} + x \cdot x = \frac{4}{3}

x, x^{2} y = 125

What are the solutions to the equation t^2+2t+6=7t+2 ?
The solutions to the equation t^2+2t+6=7t+2 are t=4,t=1
Find the zeros of t^2+2t+6=7t+2
The zeros of t^2+2t+6=7t+2 are t=4,t=1