What are the solutions to the equation 2x^2+25=13x ?

The solutions to the equation 2x^2+25=13x are x= 13/4+i(sqrt(31))/4 ,x= 13/4-i(sqrt(31))/4

Find the zeros of 2x^2+25=13x

The zeros of 2x^2+25=13x are x= 13/4+i(sqrt(31))/4 ,x= 13/4-i(sqrt(31))/4

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2 x^{2} + 25 = 13 x

x = \frac{13}{4} + i \frac{31}{4}, x = \frac{13}{4} - i \frac{31}{4}

Solution steps

2 x^{2} + 25 = 13 x

Move

13 x

to the left side

2 x^{2} + 25 - 13 x = 0

Write in the standard form

a x^{2} + b x + c = 0

2 x^{2} - 13 x + 25 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 13 ) \pm ( - 13 ) ^{2} - 4 \cdot 2 \cdot 25}{2 \cdot 2}

Simplify

(- 13)^{2} - 4 \cdot 2 \cdot 25 : 31 i

x_{1, 2} = \frac{- ( - 13 ) \pm 31 i}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 13 ) + 31 i}{2 \cdot 2}, x_{2} = \frac{- ( - 13 ) - 31 i}{2 \cdot 2}

x = \frac{- ( - 13 ) + 31 i}{2 \cdot 2} : \frac{13}{4} + i \frac{31}{4}

x = \frac{- ( - 13 ) - 31 i}{2 \cdot 2} : \frac{13}{4} - i \frac{31}{4}

The solutions to the quadratic equation are:

x = \frac{13}{4} + i \frac{31}{4}, x = \frac{13}{4} - i \frac{31}{4}

5 y^{2} - 21 y + 18

\frac{m + 7}{m ^{2} + 4 m - 21}

∣ x ∣ - ∣ x - 1 ∣ + ∣ x + 2 ∣ \geq 0

6 \cdot 15

(5 + 7)^{2}

What are the solutions to the equation 2x^2+25=13x ?
The solutions to the equation 2x^2+25=13x are x= 13/4+i(sqrt(31))/4 ,x= 13/4-i(sqrt(31))/4
Find the zeros of 2x^2+25=13x
The zeros of 2x^2+25=13x are x= 13/4+i(sqrt(31))/4 ,x= 13/4-i(sqrt(31))/4