Final answer:
To solve these quadratic equations, we can use the quadratic formula. The solutions for the given equations are: a) x = 7, -4; b) x = -1.37, 0.87; c) x = 1 + √3, 1 - √3; d) x = -4, 0.
Step-by-step explanation:
To solve these quadratic equations, we can use the quadratic formula.
a) For the equation x^2-12x-28=0, the quadratic formula gives us: x = (-b ± √(b^2 - 4ac)) / (2a). Plugging in the values a=1, b=-12, and c=-28, we get x = (12 ± √(144 - 4(1)(-28))) / (2(1)). Simplifying gives us x = (12 ± √(400)) / 2. The two solutions are x = 7 and x = -4.
b) For the equation 2x^2+4x-5=0, the quadratic formula gives us: x = (-b ± √(b^2 - 4ac)) / (2a). Plugging in the values a=2, b=4, and c=-5, we get x = (-4 ± √(16 - 4(2)(-5))) / (2(2)). Simplifying gives us x = (-4 ± √(56)) / 4. The two solutions are x = -1.37 and x = 0.87.
c) For the equation x^2-2x=2, we first rearrange it to x^2-2x-2=0. Then, we use the quadratic formula: x = (-b ± √(b^2 - 4ac)) / (2a). Plugging in the values a=1, b=-2, and c=-2, we get x = (2 ± √(4 - 4(1)(-2))) / (2(1)). Simplifying gives us x = (2 ± √(12)) / 2. The two solutions are x = 1 + √3 and x = 1 - √3.
d) For the equation x^2+8x=0, we first rearrange it to x^2+8x+0=0. Then, we use the quadratic formula: x = (-b ± √(b^2 - 4ac)) / (2a). Plugging in the values a=1, b=8, and c=0, we get x = (-8 ± √(64 - 4(1)(0))) / (2(1)). Simplifying gives us x = (-8 ± √(64)) / 2. The two solutions are x = -4 and x = 0.