If it's not what You are looking for type in the equation solver your own equation and let us solve it.
(F)=F^2+12F+29
We move all terms to the left:
(F)-(F^2+12F+29)=0
We get rid of parentheses
-F^2+F-12F-29=0
We add all the numbers together, and all the variables
-1F^2-11F-29=0
a = -1; b = -11; c = -29;
Δ = b2-4ac
Δ = -112-4·(-1)·(-29)
Δ = 5
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-\sqrt{5}}{2*-1}=\frac{11-\sqrt{5}}{-2} $$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+\sqrt{5}}{2*-1}=\frac{11+\sqrt{5}}{-2} $
| p+8=31 | | Y=3.8x+1 | | 8+(7x-5)=17 | | Y=3.6x+1 | | x-29=-172 | | X+(x-1)-(x-1)=4 | | 5x-(6-4x)=12 | | $35.72=7.6h | | -1785=t-(-836) | | d=87+9 | | t=-44+49 | | 12x+17=-1+4x | | 15c-10-7c=40-2c(3) | | 7+6c=25 | | 1x+4=13-1x | | 4+1x=-13+1x | | b/5-3=1 | | 12x=x+40 | | 9(h-93)=9 | | 9(h−93)=9 | | h/2π=4.3 | | b4− 2=2 | | -7u+23=-4(u+1) | | 0.7m=0.84 | | 3s-4(26-s)=29 | | 4(v-3)=2v-26 | | 10/4=5n | | 2n-3/4=(10)1/4 | | H(f)=2.59(40)+47.24 | | 2x-3=39-2x | | 80-1x=8x-1 | | 59=4y-9 |