X^2+4x+5=0
3x^2-8x+13=0
5x^2-12x+9=0

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вопрос

It appears that all three equations are quadratic equations in the form ax^2 + bx + c = 0.
To solve each of them, we can use the quadratic formula: x = (-b ± √(b^2 - 4ac)) / 2a.

1) For x^2 + 4x + 5 = 0:
a = 1, b = 4, c = 5.
Using the formula, x = (-4 ± √(4^2 - 415)) / (2*1)
x = (-4 ± √(16 - 20)) / 2
x = (-4 ± √(-4)) / 2
x = (-4 ± 2i) / 2
x = -2 ± i.

Therefore, the solutions are x = -2 + i and x = -2 - i.

2) For 3x^2 - 8x + 13 = 0:
a = 3, b = -8, c = 13.
Using the formula, x = (8 ± √((-8)^2 - 4313)) / (2*3)
x = (8 ± √(64 - 156)) / 6
x = (8 ± √(-92)) / 6
x = (8 ± 2√23 i) / 6
x = (4 ± √23 i) / 3.

Therefore, the solutions are x = (4 + √23 i) / 3 and x = (4 - √23 i) / 3.

3) For 5x^2 - 12x + 9 = 0:
a = 5, b = -12, c = 9.
Using the formula, x = (12 ± √((-12)^2 - 459)) / 2*5
x = (12 ± √(144 - 180))/10
x = (12 ± √(-36))/10
x = (12 ± 6i)/10
x = (6 ± 3i)/5.

Therefore, the solutions are x = (6 + 3i)/5 and x = (6 - 3i)/5.