What are the solutions to the equation (3x)^2-4(x+5)(x)=0 ?

The solutions to the equation (3x)^2-4(x+5)(x)=0 are x=4,x=0

Find the zeros of (3x)^2-4(x+5)(x)=0

The zeros of (3x)^2-4(x+5)(x)=0 are x=4,x=0

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(3 x)^{2} - 4 (x + 5) (x) = 0

x = 4, x = 0

Solution steps

(3 x)^{2} - 4 (x + 5) (x) = 0

Expand

(3 x)^{2} - 4 (x + 5) (x) : 5 x^{2} - 20 x

5 x^{2} - 20 x = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 20 ) \pm ( - 20 ) ^{2} - 4 \cdot 5 \cdot 0}{2 \cdot 5}

(- 20)^{2} - 4 \cdot 5 \cdot 0 = 20

x_{1, 2} = \frac{- ( - 20 ) \pm 20}{2 \cdot 5}

Separate the solutions

x_{1} = \frac{- ( - 20 ) + 20}{2 \cdot 5}, x_{2} = \frac{- ( - 20 ) - 20}{2 \cdot 5}

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

x = \frac{- ( - 20 ) - 20}{2 \cdot 5} : 0

The solutions to the quadratic equation are:

x = 4, x = 0

x, y^{3} - x y + x^{2} = 1

0.01 x^{2} - 3 x + 229 = 429 - 2 x

u^{2} + 2 u - 1 = 0

- 7 x^{2} + 4 x + 6 = 0

z^{2} - 7 z - 8 = 0

What are the solutions to the equation (3x)^2-4(x+5)(x)=0 ?
The solutions to the equation (3x)^2-4(x+5)(x)=0 are x=4,x=0
Find the zeros of (3x)^2-4(x+5)(x)=0
The zeros of (3x)^2-4(x+5)(x)=0 are x=4,x=0