If it's not what You are looking for type in the equation solver your own equation and let us solve it.
24x^2+19x-59=0
a = 24; b = 19; c = -59;
Δ = b2-4ac
Δ = 192-4·24·(-59)
Δ = 6025
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{6025}=\sqrt{25*241}=\sqrt{25}*\sqrt{241}=5\sqrt{241}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-5\sqrt{241}}{2*24}=\frac{-19-5\sqrt{241}}{48} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+5\sqrt{241}}{2*24}=\frac{-19+5\sqrt{241}}{48} $
| (3x-5)(2x+7)=15 | | 4x/x=64 | | 13x+8=14x+2x= | | x2-7x+20=0 | | 13x+8=14x+2= | | 5(2x-1)=12x | | 3x+7x+x=121 | | x/2+2x+3/3=1 | | x/x+10=0.04 | | t-(2t+5)(1-2t)=2(3+4t)-3(t-4)t= | | t-(2t+5)(1-2t)=2(3+4t)-3(t-4) | | 5^x-1=1/25 | | f(1)=5(1)2−3√1+2 | | 5÷x-4=20 | | (x+3)(x+4)=42 | | 3x^-x=10 | | -4.9x^2-214x+1.7=0 | | 5*(4x+7)-23=63 | | √3x=36 | | m^2+6m-2=0 | | 3x√-36=0 | | 3x/5=15/5 | | 5^x-2=1/25 | | 3x2-8=67 | | 3/x=1/9 | | (x−6)(x2−5x+16)=0 | | 4x-12+10=3x-x+6 | | X^2-21x=100 | | 4.5+x=9/4x | | 2y-10y+15=2y+5 | | 3w-14+5w=6+8w-15 | | 3(2a+5-4+3a-1(=10 |