Solve the equation. 15x^2 – 28x + 5 = 0

Given the quadratic equation:

15x^2 - 28x + 5

a = 15

b = -28

c = 5

Solve for x.

Solution:

In order to solve for x, we simply must look at the 2nd and 3rd value. First we must think of two numbers that when added, the answer is -28, and when multiplied, the answer is 5. Hence, if we are unable to find the number, we will use the quadratic formula.

The Quadratic formula:

x = −b ± √(b^2 − 4ac)/2a

is used to solve quadratic equations where a ≠ 0, in the form
ax^2+bx+c=0

When b^2−4ac=0 there is one real root.

When b^2−4ac>0 there are two real roots.

When b^2−4ac<0 there are no real roots, only a complex number.

Substitute the given values of a, b, and c to the quadratic formula.

x = −b ± √(b^2 − 4ac)/2a
x = −(-28) ± √((-28)^2 − 4(15)(5))/2(15)
x = 28 ± √(784 − 300)/30
x = 28 ± √(484)/30
x = 28 ± 22/30

Solve for + - separately.

x = 28 + 22 / 30
x = 50/30
x = 5/3

x = 28 - 22 / 30
x = 6/30
x = 1/5

Final answer:

x = 5/3

x = 1/5