The document provides work for solving two quadratic equations. For the first equation, x2-4x-15=0, the steps show isolating the x2 term and solving the quadratic formula to find the solutions as x=6.355 and x=2.355. For the second equation, x2-4x-12=0, the same process of isolating the x2 term and using the quadratic formula results in solutions of x=6 and x=-2.

1. Cindy Giroux
Week 3 Project 1&2
Project 1
X2-4x-15=0 4(2square)=x4=16
+15 +15
A. X2-4X=15
B. 4X2(SQUARE)-16X=60
C. +16 +16
4X2(SQUARE)-16X+16=76
D. Take square root of both sides
Square root of 4=2 2x-4=8.71
Square root of 16=4
Square root of 76=8.71
2x-4=8.71 2x-4=-8.71
+4 +4+4 +4
2x=12.71 2x = -4.71
2Divided by 2=1x 2 divided by 2
X=6.355 X=2.355

2. Second part of project 1
X2(square)-4x-12=0 4x2(square)=16
X2-4x=12
Multiply each by 4
4x2(square)-16x=48
+16 +16
4x2(square)-16x+16=64 Square Root of 64=8
2x -4 = 8
2x – 4 = 8 2x – 4 = -8
+4 +4+4 +4
2x =12 Divide by 2 2x = 4
X=6 X=-2

4. E.