What are the solutions to the equation (3x+1)^2-(x+4)^2=-15 ?

The solutions to the equation (3x+1)^2-(x+4)^2=-15 are x= 1/4 ,x=0

Find the zeros of (3x+1)^2-(x+4)^2=-15

The zeros of (3x+1)^2-(x+4)^2=-15 are x= 1/4 ,x=0

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(3 x + 1)^{2} - (x + 4)^{2} = - 15

x = \frac{1}{4}, x = 0

Decimal Notation

x = 0.25, x = 0

Solution steps

(3 x + 1)^{2} - (x + 4)^{2} = - 15

Expand

(3 x + 1)^{2} - (x + 4)^{2} : 8 x^{2} - 2 x - 15

8 x^{2} - 2 x - 15 = - 15

Move

15

to the left side

8 x^{2} - 2 x = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 8 \cdot 0}{2 \cdot 8}

(- 2)^{2} - 4 \cdot 8 \cdot 0 = 2

x_{1, 2} = \frac{- ( - 2 ) \pm 2}{2 \cdot 8}

Separate the solutions

x_{1} = \frac{- ( - 2 ) + 2}{2 \cdot 8}, x_{2} = \frac{- ( - 2 ) - 2}{2 \cdot 8}

x = \frac{- ( - 2 ) + 2}{2 \cdot 8} : \frac{1}{4}

x = \frac{- ( - 2 ) - 2}{2 \cdot 8} : 0

The solutions to the quadratic equation are:

x = \frac{1}{4}, x = 0

(x + 2)^{2} - 6 = 4 x - 2

p^{2} + p = 0

4 t = t^{2}

k^{2} - 9 k + 20 = 0

x^{2} - 2.455 x + 1.25 = 0

What are the solutions to the equation (3x+1)^2-(x+4)^2=-15 ?
The solutions to the equation (3x+1)^2-(x+4)^2=-15 are x= 1/4 ,x=0
Find the zeros of (3x+1)^2-(x+4)^2=-15
The zeros of (3x+1)^2-(x+4)^2=-15 are x= 1/4 ,x=0