What are the solutions to the equation (x-4)^2+3(x+1)(x-2)=8 ?

The solutions to the equation (x-4)^2+3(x+1)(x-2)=8 are x=(11+sqrt(89))/8 ,x=(11-sqrt(89))/8

Find the zeros of (x-4)^2+3(x+1)(x-2)=8

The zeros of (x-4)^2+3(x+1)(x-2)=8 are x=(11+sqrt(89))/8 ,x=(11-sqrt(89))/8

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

x = \frac{11 + 89}{8}, x = \frac{11 - 89}{8}

Solution steps

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

Expand

(x - 4)^{2} + 3 (x + 1) (x - 2) : 4 x^{2} - 11 x + 10

4 x^{2} - 11 x + 10 = 8

Move

8

to the left side

4 x^{2} - 11 x + 2 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 11 ) \pm ( - 11 ) ^{2} - 4 \cdot 4 \cdot 2}{2 \cdot 4}

(- 11)^{2} - 4 \cdot 4 \cdot 2 = 89

x_{1, 2} = \frac{- ( - 11 ) \pm 89}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 11 ) + 89}{2 \cdot 4}, x_{2} = \frac{- ( - 11 ) - 89}{2 \cdot 4}

x = \frac{- ( - 11 ) + 89}{2 \cdot 4} : \frac{11 + 89}{8}

x = \frac{- ( - 11 ) - 89}{2 \cdot 4} : \frac{11 - 89}{8}

The solutions to the quadratic equation are:

x = \frac{11 + 89}{8}, x = \frac{11 - 89}{8}

x, f (x + 2) = x^{2} - x

r^{2} - 144 = 0

4 x^{2} + 26 x + 30 = 0

- 5 x^{2} + 10 = 0

15 x^{2} = 540

What are the solutions to the equation (x-4)^2+3(x+1)(x-2)=8 ?
The solutions to the equation (x-4)^2+3(x+1)(x-2)=8 are x=(11+sqrt(89))/8 ,x=(11-sqrt(89))/8
Find the zeros of (x-4)^2+3(x+1)(x-2)=8
The zeros of (x-4)^2+3(x+1)(x-2)=8 are x=(11+sqrt(89))/8 ,x=(11-sqrt(89))/8