Question

How do you solve `7(x^2-2x+3) = -3(x^2-5x)`?

Answer 1

Let's do the multiplications at both sides:
`7x^2-14x+21=-3x^2+15x`
Add `3x^2` at both sides:
`10x^2-14x+21=15x`
Subtract `15x` from both sides:
`10x^2-29x+21=0`

So we have a quadratic equation of the form `ax^2+bx+c`, where `a=10`, `b=-29`, and `c=21`. Let's calculate the discriminant:
`\Delta=b^2-4ac= (-29^2)-4*10*21=841-840=1`
Since `\Delta>0`, there are two solutions `x_1` and `x_2`, given by
`x_{1,2}=\frac{-b\pm \sqrt(\Delta)}{2a}`
Since `\Delta=1`, its square root is also 1. The two solutions are thus
`\frac{29\pm1}{20}`
So, `x_1={29+1}/{20}=3/2` is the first solution, while `x_2={29-1}/{20}=7/5` is the second.

Answer 2

Have a look: