What are the solutions to the equation 4x^2-19x=-25 ?

The solutions to the equation 4x^2-19x=-25 are x= 19/8+i(sqrt(39))/8 ,x= 19/8-i(sqrt(39))/8

Find the zeros of 4x^2-19x=-25

The zeros of 4x^2-19x=-25 are x= 19/8+i(sqrt(39))/8 ,x= 19/8-i(sqrt(39))/8

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4 x^{2} - 19 x = - 25

x = \frac{19}{8} + i \frac{39}{8}, x = \frac{19}{8} - i \frac{39}{8}

Solution steps

4 x^{2} - 19 x = - 25

Move

25

to the left side

4 x^{2} - 19 x + 25 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 19 ) \pm ( - 19 ) ^{2} - 4 \cdot 4 \cdot 25}{2 \cdot 4}

Simplify

(- 19)^{2} - 4 \cdot 4 \cdot 25 : 39 i

x_{1, 2} = \frac{- ( - 19 ) \pm 39 i}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 19 ) + 39 i}{2 \cdot 4}, x_{2} = \frac{- ( - 19 ) - 39 i}{2 \cdot 4}

x = \frac{- ( - 19 ) + 39 i}{2 \cdot 4} : \frac{19}{8} + i \frac{39}{8}

x = \frac{- ( - 19 ) - 39 i}{2 \cdot 4} : \frac{19}{8} - i \frac{39}{8}

The solutions to the quadratic equation are:

x = \frac{19}{8} + i \frac{39}{8}, x = \frac{19}{8} - i \frac{39}{8}

7 - 3 x^{2} = 211

∣ 3 a + 1 ∣ = ∣ 2 a - 4 ∣

2 \cdot 53

6 x^{2} + 5 x - 21

q^{2} = - 1

What are the solutions to the equation 4x^2-19x=-25 ?
The solutions to the equation 4x^2-19x=-25 are x= 19/8+i(sqrt(39))/8 ,x= 19/8-i(sqrt(39))/8
Find the zeros of 4x^2-19x=-25
The zeros of 4x^2-19x=-25 are x= 19/8+i(sqrt(39))/8 ,x= 19/8-i(sqrt(39))/8