What are the solutions to the equation (-t)^2-4*3*(-2)=1 ?

The solutions to the equation (-t)^2-4*3*(-2)=1 are t=-sqrt(23)i,t=sqrt(23)i

Find the zeros of (-t)^2-4*3*(-2)=1

The zeros of (-t)^2-4*3*(-2)=1 are t=-sqrt(23)i,t=sqrt(23)i

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(- t)^{2} - 4 \cdot 3 \cdot (- 2) = 1

t = - 23 i, t = 23 i

Solution steps

(- t)^{2} - 4 \cdot 3 (- 2) = 1

Move

4 \cdot 3 \cdot (- 2)

to the right side

(- t)^{2} = - 23

For

(g (x))^{2} = f (a)

the solutions are

g (x) = f (a), - f (a)

Solve

- t = - 23 : t = - 23 i

Solve

- t = - - 23 : t = 23 i

The solutions to the quadratic equation are:

t = - 23 i, t = 23 i

x, 3 x^{2} + 2 y^{2} + e^{5 y} = 13

x^{2} + 105 x - 2500 = 0

x^{2} = 15 + 2 x

r^{2} + 120 r + 10000 = 0

x (x + 3) - 5 (2 x - 4) = 2 x

What are the solutions to the equation (-t)^2-4*3*(-2)=1 ?
The solutions to the equation (-t)^2-4*3*(-2)=1 are t=-sqrt(23)i,t=sqrt(23)i
Find the zeros of (-t)^2-4*3*(-2)=1
The zeros of (-t)^2-4*3*(-2)=1 are t=-sqrt(23)i,t=sqrt(23)i