What are the solutions to the equation 4x^2-20x=-29 ?

The solutions to the equation 4x^2-20x=-29 are x= 5/2+i,x= 5/2-i

Find the zeros of 4x^2-20x=-29

The zeros of 4x^2-20x=-29 are x= 5/2+i,x= 5/2-i

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4 x^{2} - 20 x = - 29

x = \frac{5}{2} + i, x = \frac{5}{2} - i

Solution steps

4 x^{2} - 20 x = - 29

Move

29

to the left side

4 x^{2} - 20 x + 29 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 20 ) \pm ( - 20 ) ^{2} - 4 \cdot 4 \cdot 29}{2 \cdot 4}

Simplify

(- 20)^{2} - 4 \cdot 4 \cdot 29 : 8 i

x_{1, 2} = \frac{- ( - 20 ) \pm 8 i}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 20 ) + 8 i}{2 \cdot 4}, x_{2} = \frac{- ( - 20 ) - 8 i}{2 \cdot 4}

x = \frac{- ( - 20 ) + 8 i}{2 \cdot 4} : \frac{5}{2} + i

x = \frac{- ( - 20 ) - 8 i}{2 \cdot 4} : \frac{5}{2} - i

The solutions to the quadratic equation are:

x = \frac{5}{2} + i, x = \frac{5}{2} - i

5 (x - 9)^{2} = 20

y, x^{3} + 4 x^{2} + x y^{2} - 4 y^{2} = 0

3 y^{2} - 11 y + 10 = 0

x, x^{2} + y^{2} + 3 x + y + 10 = 0

a, a^{2} + 7^{2} = 1 6^{2}

What are the solutions to the equation 4x^2-20x=-29 ?
The solutions to the equation 4x^2-20x=-29 are x= 5/2+i,x= 5/2-i
Find the zeros of 4x^2-20x=-29
The zeros of 4x^2-20x=-29 are x= 5/2+i,x= 5/2-i