Name:                                    Date:

Solve the following quadratic equations:

Question 1: ${\Large\frac{48e+49}{2e+7}}=5e-2$	Question 2: ${\Large\frac{47t-40}{7t+2}}=3t-2$
Question 3: $75c^2-19c=-6c+3c^2+20$	Question 4: $-{\Large\frac{77}{j+2}}+{\Large\frac{180}{j+3}}=20$
Question 5: ${\Large\frac{-123a-78}{5a-3}}=8a+2$	Question 6: ${\Large\frac{8}{z-2}}-{\Large\frac{15}{z-1}}=6$
Question 7: ${\Large\frac{-42v+19}{9v-8}}=7v+2$	Question 8: $-{\Large\frac{35}{m+1}}+{\Large\frac{84}{m+2}}=10$
Question 9: $(s-3)^2+(2s-7)^2=(2s-5)^2$	Question 10: $(q+6)^2+(q+7)^2=(q+8)^2$
Question 11: $(w+6)^2+(2w+14)^2=(3w+16)^2$	Question 12: $(h+2)^2+(3h+9)^2=(4h+5)^2$
Question 13: $3d^2-49d=-2d-12d^2-36$	Question 14: $38n^2-57n=-5n+6n^2-15$
Question 15: $-{\Large\frac{15}{g}}+{\Large\frac{45}{g+1}}=8$	Question 16: $13b^2-21b=-12b-5b^2+20$

Answers

Question 1: $(2e-7)(5e+9)=0$ $\implies e={\Large\frac{7}{2}} , e=-{\Large\frac{9}{5}}$	Question 2: $(7t-9)(3t-4)=0$ $\implies t={\Large\frac{9}{7}} , t={\Large\frac{4}{3}}$
Question 3: $(9c+4)(8c-5)=0$ $\implies c=-{\Large\frac{4}{9}} , c={\Large\frac{5}{8}}$	Question 4: $(4j-3)(5j+3)=0$ $\implies j={\Large\frac{3}{4}} , j=-{\Large\frac{3}{5}}$
Question 5: $(5a+8)(8a+9)=0$ $\implies a=-{\Large\frac{8}{5}} , a=-{\Large\frac{9}{8}}$	Question 6: $(3z+2)(2z-5)=0$ $\implies z=-{\Large\frac{2}{3}} , z={\Large\frac{5}{2}}$
Question 7: $(9v+7)(7v-5)=0$ $\implies v=-{\Large\frac{7}{9}} , v={\Large\frac{5}{7}}$	Question 8: $(5m-2)(2m-3)=0$ $\implies m={\Large\frac{2}{5}} , m={\Large\frac{3}{2}}$
Question 9: $(s-3)(s-11)=0$ $\implies s=3 , s=11$	Question 10: $(q+3)(q+7)=0$ $\implies q=-3 , q=-7$
Question 11: $4(w+6)(w+1)=0$ $\implies w=-6 , w=-1$	Question 12: $6(h+2)(h-5)=0$ $\implies h=-2 , h=5$
Question 13: $(3d-4)(5d-9)=0$ $\implies d={\Large\frac{4}{3}} , d={\Large\frac{9}{5}}$	Question 14: $(8n-3)(4n-5)=0$ $\implies n={\Large\frac{3}{8}} , n={\Large\frac{5}{4}}$
Question 15: $(2g-3)(4g-5)=0$ $\implies g={\Large\frac{3}{2}} , g={\Large\frac{5}{4}}$	Question 16: $(6b+5)(3b-4)=0$ $\implies b=-{\Large\frac{5}{6}} , b={\Large\frac{4}{3}}$