What are the solutions to the equation t^2+10t=25 ?

The solutions to the equation t^2+10t=25 are t=5(sqrt(2)-1),t=-5(1+sqrt(2))

Find the zeros of t^2+10t=25

The zeros of t^2+10t=25 are t=5(sqrt(2)-1),t=-5(1+sqrt(2))

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t^{2} + 10 t = 25

t = 5 (2 - 1), t = - 5 (1 + 2)

Decimal Notation

t = 2.07106 \dots, t = - 12.07106 \dots

Solution steps

t^{2} + 10 t = 25

Move

25

to the left side

t^{2} + 10 t - 25 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 10 \pm 1 0 ^{2} - 4 \cdot 1 \cdot ( - 25 )}{2 \cdot 1}

1 0^{2} - 4 \cdot 1 \cdot (- 25) = 102

t_{1, 2} = \frac{- 10 \pm 10 2}{2 \cdot 1}

Separate the solutions

t_{1} = \frac{- 10 + 10 2}{2 \cdot 1}, t_{2} = \frac{- 10 - 10 2}{2 \cdot 1}

t = \frac{- 10 + 10 2}{2 \cdot 1} : 5 (2 - 1)

t = \frac{- 10 - 10 2}{2 \cdot 1} : - 5 (1 + 2)

The solutions to the quadratic equation are:

t = 5 (2 - 1), t = - 5 (1 + 2)

1. 2^{3}

(3 s t^{12})^{3}

4 x + 5 \leq 10

1 + 3 s \geq 10

∣ 5 x + 1 ∣ = ∣ 8 x + 1 ∣

What are the solutions to the equation t^2+10t=25 ?
The solutions to the equation t^2+10t=25 are t=5(sqrt(2)-1),t=-5(1+sqrt(2))
Find the zeros of t^2+10t=25
The zeros of t^2+10t=25 are t=5(sqrt(2)-1),t=-5(1+sqrt(2))