What are the solutions to the equation (2x+1)^2-3(2x+1)-10=0 ?

The solutions to the equation (2x+1)^2-3(2x+1)-10=0 are x=2,x=-3/2

Find the zeros of (2x+1)^2-3(2x+1)-10=0

The zeros of (2x+1)^2-3(2x+1)-10=0 are x=2,x=-3/2

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(2 x + 1)^{2} - 3 (2 x + 1) - 10 = 0

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

Decimal Notation

x = 2, x = - 1.5

Solution steps

(2 x + 1)^{2} - 3 (2 x + 1) - 10 = 0

Expand

(2 x + 1)^{2} - 3 (2 x + 1) - 10 : 4 x^{2} - 2 x - 12

4 x^{2} - 2 x - 12 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 4 ( - 12 )}{2 \cdot 4}

(- 2)^{2} - 4 \cdot 4 (- 12) = 14

x_{1, 2} = \frac{- ( - 2 ) \pm 14}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 2 ) + 14}{2 \cdot 4}, x_{2} = \frac{- ( - 2 ) - 14}{2 \cdot 4}

x = \frac{- ( - 2 ) + 14}{2 \cdot 4} : 2

x = \frac{- ( - 2 ) - 14}{2 \cdot 4} : - \frac{3}{2}

The solutions to the quadratic equation are:

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

\frac{4}{9} = \frac{r - 3}{6}

x - \frac{1}{3} \leq \frac{7}{12}

\frac{7}{3} - \frac{32}{3}

0 = - 0.6 (25)^{2} + 30 (25)

\frac{1 + 2}{2 ( 1 ) + 6}

What are the solutions to the equation (2x+1)^2-3(2x+1)-10=0 ?
The solutions to the equation (2x+1)^2-3(2x+1)-10=0 are x=2,x=-3/2
Find the zeros of (2x+1)^2-3(2x+1)-10=0
The zeros of (2x+1)^2-3(2x+1)-10=0 are x=2,x=-3/2