What are the solutions to the equation (2x+1)^2-10x=19 ?

The solutions to the equation (2x+1)^2-10x=19 are x=3,x=-3/2

Find the zeros of (2x+1)^2-10x=19

The zeros of (2x+1)^2-10x=19 are x=3,x=-3/2

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(2 x + 1)^{2} - 10 x = 19

x = 3, x = - \frac{3}{2}

Decimal Notation

x = 3, x = - 1.5

Solution steps

(2 x + 1)^{2} - 10 x = 19

Expand

(2 x + 1)^{2} - 10 x : 4 x^{2} - 6 x + 1

4 x^{2} - 6 x + 1 = 19

Move

19

to the left side

4 x^{2} - 6 x - 18 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 \cdot 4 ( - 18 )}{2 \cdot 4}

(- 6)^{2} - 4 \cdot 4 (- 18) = 18

x_{1, 2} = \frac{- ( - 6 ) \pm 18}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 6 ) + 18}{2 \cdot 4}, x_{2} = \frac{- ( - 6 ) - 18}{2 \cdot 4}

x = \frac{- ( - 6 ) + 18}{2 \cdot 4} : 3

x = \frac{- ( - 6 ) - 18}{2 \cdot 4} : - \frac{3}{2}

The solutions to the quadratic equation are:

x = 3, x = - \frac{3}{2}

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

a, a^{2} - 2 ab + b^{2} = 1

2 g^{2} + 8 g + 8 = 0

x^{2} - 10 = - 6 x

x^{2} - 10 = - 10

What are the solutions to the equation (2x+1)^2-10x=19 ?
The solutions to the equation (2x+1)^2-10x=19 are x=3,x=-3/2
Find the zeros of (2x+1)^2-10x=19
The zeros of (2x+1)^2-10x=19 are x=3,x=-3/2