What are the solutions to the equation (4+3x)^2+x^2=146 ?

The solutions to the equation (4+3x)^2+x^2=146 are x= 13/5 ,x=-5

Find the zeros of (4+3x)^2+x^2=146

The zeros of (4+3x)^2+x^2=146 are x= 13/5 ,x=-5

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(4 + 3 x)^{2} + x^{2} = 146

x = \frac{13}{5}, x = - 5

Decimal Notation

x = 2.6, x = - 5

Solution steps

(4 + 3 x)^{2} + x^{2} = 146

Expand

(4 + 3 x)^{2} + x^{2} : 16 + 24 x + 10 x^{2}

16 + 24 x + 10 x^{2} = 146

Move

146

to the left side

10 x^{2} + 24 x - 130 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 24 \pm 2 4 ^{2} - 4 \cdot 10 ( - 130 )}{2 \cdot 10}

2 4^{2} - 4 \cdot 10 (- 130) = 76

x_{1, 2} = \frac{- 24 \pm 76}{2 \cdot 10}

Separate the solutions

x_{1} = \frac{- 24 + 76}{2 \cdot 10}, x_{2} = \frac{- 24 - 76}{2 \cdot 10}

x = \frac{- 24 + 76}{2 \cdot 10} : \frac{13}{5}

x = \frac{- 24 - 76}{2 \cdot 10} : - 5

The solutions to the quadratic equation are:

x = \frac{13}{5}, x = - 5

7 x^{2} + 8 x - 9 = 10

- 5 x^{2} + 11 x + 36 = 0

3 x^{2} + 5 = 59

x^{2} = 2 (x + 12)

t^{2} + 4 t + 12 = 0

What are the solutions to the equation (4+3x)^2+x^2=146 ?
The solutions to the equation (4+3x)^2+x^2=146 are x= 13/5 ,x=-5
Find the zeros of (4+3x)^2+x^2=146
The zeros of (4+3x)^2+x^2=146 are x= 13/5 ,x=-5