If it's not what You are looking for type in the equation solver your own equation and let us solve it.
9x^2+78x+1=0
a = 9; b = 78; c = +1;
Δ = b2-4ac
Δ = 782-4·9·1
Δ = 6048
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{6048}=\sqrt{144*42}=\sqrt{144}*\sqrt{42}=12\sqrt{42}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(78)-12\sqrt{42}}{2*9}=\frac{-78-12\sqrt{42}}{18} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(78)+12\sqrt{42}}{2*9}=\frac{-78+12\sqrt{42}}{18} $
| 8(4-3)+1=53-3(x-5) | | 6v=3 | | 4z/7+9=8 | | 3^n=5 | | b-1/2=3/4 | | 10=2(v-4)-8v | | 13+w/7=–18 | | 5z-9=9+7 | | 8/17=m/5 | | 0.5x²+1=9 | | 8x-x²=0 | | 0=8x-x² | | 2b/4-4=1 | | 2x/3=12/10(x×1) | | 1/(2x+1)-1/(3x+1)=3 | | 3(x-6)=-48 | | 11x-4/5=8 | | 1/2X+1-1/3x+1=3 | | (1/(2x+1))/(1/(3x+1))=3 | | x/9+9=1 | | 18=3(2y-4) | | 5(3w+1)=-9 | | 5(3w+1)=-3 | | 3^(2*x)-2^(2*x)+0.5^(1-2*x)=0 | | 3^(2*x)-2^(2*x)+(5/10)^(1-2*x)=0 | | z=8-3 | | 6r+7=20 | | (1+r)^7-8r-1=0 | | 8x^2-23x+17=-8x+3 | | 3^(2*x)-(2^(2*x)+0.5(1-2*x))=0 | | 3^(2*x)-(2^(2*x)+(5/10)^(1-2*x))=0 | | 79-3x^2-8x=5 |