What are the solutions to the equation 4x^2-8x=40 ?

The solutions to the equation 4x^2-8x=40 are x=1+sqrt(11),x=1-sqrt(11)

Find the zeros of 4x^2-8x=40

The zeros of 4x^2-8x=40 are x=1+sqrt(11),x=1-sqrt(11)

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4 x^{2} - 8 x = 40

x = 1 + 11, x = 1 - 11

Decimal Notation

x = 4.31662 \dots, x = - 2.31662 \dots

Solution steps

4 x^{2} - 8 x = 40

Move

40

to the left side

4 x^{2} - 8 x - 40 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 \cdot 4 ( - 40 )}{2 \cdot 4}

(- 8)^{2} - 4 \cdot 4 (- 40) = 811

x_{1, 2} = \frac{- ( - 8 ) \pm 8 11}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 8 ) + 8 11}{2 \cdot 4}, x_{2} = \frac{- ( - 8 ) - 8 11}{2 \cdot 4}

x = \frac{- ( - 8 ) + 8 11}{2 \cdot 4} : 1 + 11

x = \frac{- ( - 8 ) - 8 11}{2 \cdot 4} : 1 - 11

The solutions to the quadratic equation are:

x = 1 + 11, x = 1 - 11

25 x^{2} + 20 x + 4 = 0

12 = 6 y - 36

\frac{2.9}{2.9 - 3}

7 \cdot 3 - \frac{15}{5}

25 (2) + 40 (1) < 120

What are the solutions to the equation 4x^2-8x=40 ?
The solutions to the equation 4x^2-8x=40 are x=1+sqrt(11),x=1-sqrt(11)
Find the zeros of 4x^2-8x=40
The zeros of 4x^2-8x=40 are x=1+sqrt(11),x=1-sqrt(11)