What are the solutions to the equation (2x-3)^2+3(x+4)=5 ?

The solutions to the equation (2x-3)^2+3(x+4)=5 are x= 9/8+i(5sqrt(7))/8 ,x= 9/8-i(5sqrt(7))/8

Find the zeros of (2x-3)^2+3(x+4)=5

The zeros of (2x-3)^2+3(x+4)=5 are x= 9/8+i(5sqrt(7))/8 ,x= 9/8-i(5sqrt(7))/8

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(2 x - 3)^{2} + 3 (x + 4) = 5

x = \frac{9}{8} + i \frac{5 7}{8}, x = \frac{9}{8} - i \frac{5 7}{8}

Solution steps

(2 x - 3)^{2} + 3 (x + 4) = 5

Expand

(2 x - 3)^{2} + 3 (x + 4) : 4 x^{2} - 9 x + 21

4 x^{2} - 9 x + 21 = 5

Move

5

to the left side

4 x^{2} - 9 x + 16 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 9 ) \pm ( - 9 ) ^{2} - 4 \cdot 4 \cdot 16}{2 \cdot 4}

Simplify

(- 9)^{2} - 4 \cdot 4 \cdot 16 : 57 i

x_{1, 2} = \frac{- ( - 9 ) \pm 5 7 i}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 9 ) + 5 7 i}{2 \cdot 4}, x_{2} = \frac{- ( - 9 ) - 5 7 i}{2 \cdot 4}

x = \frac{- ( - 9 ) + 5 7 i}{2 \cdot 4} : \frac{9}{8} + i \frac{5 7}{8}

x = \frac{- ( - 9 ) - 5 7 i}{2 \cdot 4} : \frac{9}{8} - i \frac{5 7}{8}

The solutions to the quadratic equation are:

x = \frac{9}{8} + i \frac{5 7}{8}, x = \frac{9}{8} - i \frac{5 7}{8}

2 m^{2} + 10 m + 3 = 0

- 4.9 x^{2} + 2 x + 10 = 0

- 5 x^{2} - 5 x = 0

x^{2} - 8 x + 9 = 56

x, (x + y) (x + y) = 1

What are the solutions to the equation (2x-3)^2+3(x+4)=5 ?
The solutions to the equation (2x-3)^2+3(x+4)=5 are x= 9/8+i(5sqrt(7))/8 ,x= 9/8-i(5sqrt(7))/8
Find the zeros of (2x-3)^2+3(x+4)=5
The zeros of (2x-3)^2+3(x+4)=5 are x= 9/8+i(5sqrt(7))/8 ,x= 9/8-i(5sqrt(7))/8