What are the solutions to the equation 0.4x^2-1.2x-0.36=0 ?

The solutions to the equation 0.4x^2-1.2x-0.36=0 are x=(3(5+sqrt(35)))/(10),x=(3(5-sqrt(35)))/(10)

Find the zeros of 0.4x^2-1.2x-0.36=0

The zeros of 0.4x^2-1.2x-0.36=0 are x=(3(5+sqrt(35)))/(10),x=(3(5-sqrt(35)))/(10)

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0.4 x^{2} - 1.2 x - 0.36 = 0

x = \frac{3 ( 5 + 35 )}{10}, x = \frac{3 ( 5 - 35 )}{10}

Decimal Notation

x = 3.27482 \dots, x = - 0.27482 \dots

Solution steps

0.4 x^{2} - 1.2 x - 0.36 = 0

Multiply both sides by

100

40 x^{2} - 120 x - 36 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 120 ) \pm ( - 120 ) ^{2} - 4 \cdot 40 ( - 36 )}{2 \cdot 40}

(- 120)^{2} - 4 \cdot 40 (- 36) = 2435

x_{1, 2} = \frac{- ( - 120 ) \pm 24 35}{2 \cdot 40}

Separate the solutions

x_{1} = \frac{- ( - 120 ) + 24 35}{2 \cdot 40}, x_{2} = \frac{- ( - 120 ) - 24 35}{2 \cdot 40}

x = \frac{- ( - 120 ) + 24 35}{2 \cdot 40} : \frac{3 ( 5 + 35 )}{10}

x = \frac{- ( - 120 ) - 24 35}{2 \cdot 40} : \frac{3 ( 5 - 35 )}{10}

The solutions to the quadratic equation are:

x = \frac{3 ( 5 + 35 )}{10}, x = \frac{3 ( 5 - 35 )}{10}

(6 t + 1)^{2} + 3 = 0

0.2 x^{2} - 0.4 x - 0.2 = 0

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

36 v^{2} + 2 = 6

x (x - 2) = - 1

What are the solutions to the equation 0.4x^2-1.2x-0.36=0 ?
The solutions to the equation 0.4x^2-1.2x-0.36=0 are x=(3(5+sqrt(35)))/(10),x=(3(5-sqrt(35)))/(10)
Find the zeros of 0.4x^2-1.2x-0.36=0
The zeros of 0.4x^2-1.2x-0.36=0 are x=(3(5+sqrt(35)))/(10),x=(3(5-sqrt(35)))/(10)