What are the solutions to the equation (2x-5)^2+4x=12 ?

The solutions to the equation (2x-5)^2+4x=12 are x=(4+sqrt(3))/2 ,x=(4-sqrt(3))/2

Find the zeros of (2x-5)^2+4x=12

The zeros of (2x-5)^2+4x=12 are x=(4+sqrt(3))/2 ,x=(4-sqrt(3))/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(2 x - 5)^{2} + 4 x = 12

x = \frac{4 + 3}{2}, x = \frac{4 - 3}{2}

Decimal Notation

x = 2.86602 \dots, x = 1.13397 \dots

Solution steps

(2 x - 5)^{2} + 4 x = 12

Expand

(2 x - 5)^{2} + 4 x : 4 x^{2} - 16 x + 25

4 x^{2} - 16 x + 25 = 12

Move

12

to the left side

4 x^{2} - 16 x + 13 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 16 ) \pm ( - 16 ) ^{2} - 4 \cdot 4 \cdot 13}{2 \cdot 4}

(- 16)^{2} - 4 \cdot 4 \cdot 13 = 43

x_{1, 2} = \frac{- ( - 16 ) \pm 4 3}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 16 ) + 4 3}{2 \cdot 4}, x_{2} = \frac{- ( - 16 ) - 4 3}{2 \cdot 4}

x = \frac{- ( - 16 ) + 4 3}{2 \cdot 4} : \frac{4 + 3}{2}

x = \frac{- ( - 16 ) - 4 3}{2 \cdot 4} : \frac{4 - 3}{2}

The solutions to the quadratic equation are:

x = \frac{4 + 3}{2}, x = \frac{4 - 3}{2}

5 q^{2} + 12 q + 5 = 2 q^{2}

y^{2} + 6 y + 25 = 0

x, 3 x^{2} - 8 x - 16 = 0

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

6 x^{2} - 72 x + 162 = 0

What are the solutions to the equation (2x-5)^2+4x=12 ?
The solutions to the equation (2x-5)^2+4x=12 are x=(4+sqrt(3))/2 ,x=(4-sqrt(3))/2
Find the zeros of (2x-5)^2+4x=12
The zeros of (2x-5)^2+4x=12 are x=(4+sqrt(3))/2 ,x=(4-sqrt(3))/2