What are the solutions to the equation 4x^2-20x=-13 ?

The solutions to the equation 4x^2-20x=-13 are x=(5+2sqrt(3))/2 ,x=(5-2sqrt(3))/2

Find the zeros of 4x^2-20x=-13

The zeros of 4x^2-20x=-13 are x=(5+2sqrt(3))/2 ,x=(5-2sqrt(3))/2

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4 x^{2} - 20 x = - 13

x = \frac{5 + 2 3}{2}, x = \frac{5 - 2 3}{2}

Decimal Notation

x = 4.23205 \dots, x = 0.76794 \dots

Solution steps

4 x^{2} - 20 x = - 13

Move

13

to the left side

4 x^{2} - 20 x + 13 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 20 ) \pm ( - 20 ) ^{2} - 4 \cdot 4 \cdot 13}{2 \cdot 4}

(- 20)^{2} - 4 \cdot 4 \cdot 13 = 83

x_{1, 2} = \frac{- ( - 20 ) \pm 8 3}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 20 ) + 8 3}{2 \cdot 4}, x_{2} = \frac{- ( - 20 ) - 8 3}{2 \cdot 4}

x = \frac{- ( - 20 ) + 8 3}{2 \cdot 4} : \frac{5 + 2 3}{2}

x = \frac{- ( - 20 ) - 8 3}{2 \cdot 4} : \frac{5 - 2 3}{2}

The solutions to the quadratic equation are:

x = \frac{5 + 2 3}{2}, x = \frac{5 - 2 3}{2}

\frac{x + 1}{4} = \frac{1}{6} + \frac{2 - x}{3}

4 x - 40 < - 6 \cdot (2 - x) - 6 x

x^{2} > 2 x + 3

2 + \frac{1}{a + 5} - \frac{2 a}{a - 5}

81 x^{4} - 16 y^{4}

What are the solutions to the equation 4x^2-20x=-13 ?
The solutions to the equation 4x^2-20x=-13 are x=(5+2sqrt(3))/2 ,x=(5-2sqrt(3))/2
Find the zeros of 4x^2-20x=-13
The zeros of 4x^2-20x=-13 are x=(5+2sqrt(3))/2 ,x=(5-2sqrt(3))/2