What are the solutions to the equation 4x^2+20x-25=0 ?

The solutions to the equation 4x^2+20x-25=0 are x=(5(sqrt(2)-1))/2 ,x=-(5(1+sqrt(2)))/2

Find the zeros of 4x^2+20x-25=0

The zeros of 4x^2+20x-25=0 are x=(5(sqrt(2)-1))/2 ,x=-(5(1+sqrt(2)))/2

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

x = \frac{5 ( 2 - 1 )}{2}, x = - \frac{5 ( 1 + 2 )}{2}

Decimal Notation

x = 1.03553 \dots, x = - 6.03553 \dots

Solution steps

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

Solve with the quadratic formula

x_{1, 2} = \frac{- 20 \pm 2 0 ^{2} - 4 \cdot 4 ( - 25 )}{2 \cdot 4}

2 0^{2} - 4 \cdot 4 (- 25) = 202

x_{1, 2} = \frac{- 20 \pm 20 2}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- 20 + 20 2}{2 \cdot 4}, x_{2} = \frac{- 20 - 20 2}{2 \cdot 4}

x = \frac{- 20 + 20 2}{2 \cdot 4} : \frac{5 ( 2 - 1 )}{2}

x = \frac{- 20 - 20 2}{2 \cdot 4} : - \frac{5 ( 1 + 2 )}{2}

The solutions to the quadratic equation are:

x = \frac{5 ( 2 - 1 )}{2}, x = - \frac{5 ( 1 + 2 )}{2}

∣ (2 x) ∣ > 6

X = 1.35 \frac{( 25200 ) ( 1.125 )}{\frac{3.1416}{32} ( 2.2 5 ^{4} )}

9 w + 63 = 198

x^{2} - 9

(- 3 a^{\frac{1}{4}}) (9 a)^{- \frac{3}{2}}

What are the solutions to the equation 4x^2+20x-25=0 ?
The solutions to the equation 4x^2+20x-25=0 are x=(5(sqrt(2)-1))/2 ,x=-(5(1+sqrt(2)))/2
Find the zeros of 4x^2+20x-25=0
The zeros of 4x^2+20x-25=0 are x=(5(sqrt(2)-1))/2 ,x=-(5(1+sqrt(2)))/2