What are the solutions to the equation 1.3x^2-4x=3.5 ?

The solutions to the equation 1.3x^2-4x=3.5 are x=(20+3sqrt(95))/(13),x=(20-3sqrt(95))/(13)

Find the zeros of 1.3x^2-4x=3.5

The zeros of 1.3x^2-4x=3.5 are x=(20+3sqrt(95))/(13),x=(20-3sqrt(95))/(13)

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1.3 x^{2} - 4 x = 3.5

x = \frac{20 + 3 95}{13}, x = \frac{20 - 3 95}{13}

Decimal Notation

x = 3.78772 \dots, x = - 0.71079 \dots

Solution steps

1.3 x^{2} - 4 x = 3.5

Multiply both sides by

10

13 x^{2} - 40 x = 35

Move

35

to the left side

13 x^{2} - 40 x - 35 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 40 ) \pm ( - 40 ) ^{2} - 4 \cdot 13 ( - 35 )}{2 \cdot 13}

(- 40)^{2} - 4 \cdot 13 (- 35) = 695

x_{1, 2} = \frac{- ( - 40 ) \pm 6 95}{2 \cdot 13}

Separate the solutions

x_{1} = \frac{- ( - 40 ) + 6 95}{2 \cdot 13}, x_{2} = \frac{- ( - 40 ) - 6 95}{2 \cdot 13}

x = \frac{- ( - 40 ) + 6 95}{2 \cdot 13} : \frac{20 + 3 95}{13}

x = \frac{- ( - 40 ) - 6 95}{2 \cdot 13} : \frac{20 - 3 95}{13}

The solutions to the quadratic equation are:

x = \frac{20 + 3 95}{13}, x = \frac{20 - 3 95}{13}

x, C x^{2} + x + y^{2} = 0

x^{2} - 8 x + 16 = - 12

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

(k + 1) x^{2} - (k - 1) x - (k + 2) = 0

0 = x^{2} - 36

What are the solutions to the equation 1.3x^2-4x=3.5 ?
The solutions to the equation 1.3x^2-4x=3.5 are x=(20+3sqrt(95))/(13),x=(20-3sqrt(95))/(13)
Find the zeros of 1.3x^2-4x=3.5
The zeros of 1.3x^2-4x=3.5 are x=(20+3sqrt(95))/(13),x=(20-3sqrt(95))/(13)