What are the solutions to the equation (2t+3)^2+(t+4)^2=10 ?

The solutions to the equation (2t+3)^2+(t+4)^2=10 are t=-1,t=-3

Find the zeros of (2t+3)^2+(t+4)^2=10

The zeros of (2t+3)^2+(t+4)^2=10 are t=-1,t=-3

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(2 t + 3)^{2} + (t + 4)^{2} = 10

t = - 1, t = - 3

Solution steps

(2 t + 3)^{2} + (t + 4)^{2} = 10

Expand

(2 t + 3)^{2} + (t + 4)^{2} : 5 t^{2} + 20 t + 25

5 t^{2} + 20 t + 25 = 10

Move

10

to the left side

5 t^{2} + 20 t + 15 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 20 \pm 2 0 ^{2} - 4 \cdot 5 \cdot 15}{2 \cdot 5}

2 0^{2} - 4 \cdot 5 \cdot 15 = 10

t_{1, 2} = \frac{- 20 \pm 10}{2 \cdot 5}

Separate the solutions

t_{1} = \frac{- 20 + 10}{2 \cdot 5}, t_{2} = \frac{- 20 - 10}{2 \cdot 5}

t = \frac{- 20 + 10}{2 \cdot 5} : - 1

t = \frac{- 20 - 10}{2 \cdot 5} : - 3

The solutions to the quadratic equation are:

t = - 1, t = - 3

2 x^{2} + 2 x + 15 = 0

4 m^{2} - 32 = - 8 m

x^{2} - 8 x - 20 = - x^{2} + 10 x - 60

2 k (k + 7) = - 49

7 n^{2} - 3 n - 3 = 0

What are the solutions to the equation (2t+3)^2+(t+4)^2=10 ?
The solutions to the equation (2t+3)^2+(t+4)^2=10 are t=-1,t=-3
Find the zeros of (2t+3)^2+(t+4)^2=10
The zeros of (2t+3)^2+(t+4)^2=10 are t=-1,t=-3